User:Papaya25/sandbox

Lambert W Function
$$\int\frac{ \operatorname{W}\left(x\right) }{x} \, dx \; = \; \frac{ \operatorname{W}\left(x\right)^2}{2} + \operatorname{W}\left(x\right) + C $$

Introduce substitution variable $$ u = \operatorname{W}\left(x\right) \rightarrow ue^u=x \rightarrow \frac{dx}{du} = \left(u+1\right)e^u $$
 * $$\begin{align}

\int\frac{\operatorname{W}\left(x\right)}{x} \, dx & = \int\frac{u}{ue^u}\left(u+1\right)e^u \, du \\ & = \int\frac{\cancel{\color{OliveGreen}{u}}}{\cancel{\color{OliveGreen}{u}} \cancel{\color{BrickRed}{e^u}}}\left(u+1\right)\cancel{\color{BrickRed}{e^u}} \, du \\ & = \int u+1 \, du \\ & = \frac{u^2}{2} + u + C \\ & \quad u = \operatorname{W}\left(x\right) \\ \int\frac{\operatorname{W}\left(x\right)}{x} \, dx & = \frac{ \operatorname{W}\left(x\right) ^2}{2} + \operatorname{W}\left(x\right) + C \\ \end{align}$$


 * $$ \operatorname{W}\left(x\right) e^{ \operatorname{W}\left(x\right) } = x \rightarrow \frac{ \operatorname{W}\left(x\right) }{ x } = e^{ -\operatorname{W}\left(x\right) } $$
 * $$\begin{align}

\int\frac{\operatorname{W}\left(x\right)}{x} \, dx & = \int e^{-\operatorname{W}\left(x\right)} \, du \\ & \quad u = \operatorname{W}\left(x\right) \rightarrow ue^u=x \rightarrow \frac{dx}{du} = \left(u+1\right)e^u \\ & = \int e^{-u}\left(u+1\right)e^u \, du \\ & = \int \cancel{\color{OliveGreen}{e^{-u}}}\left(u+1\right)\cancel{\color{OliveGreen}{e^{u}}} \, du \\ & = \int u+1 \, du \\ & = \frac{u^2}{2} + u + C \\ & \quad u = \operatorname{W}\left(x\right) \\ \int\frac{\operatorname{W}\left(x\right)}{x} \, dx & = \frac{ \operatorname{W}\left(x\right) ^2}{2} + \operatorname{W}\left(x\right) + C \\ \end{align}$$

Factorial

 * $$x! = \prod_{m=1}^{\infty}\frac{m}{m+x}\left(1+\frac{1}{m}\right)^x$$


 * $$\lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x\right)!} = 1 \;\;\because$$
 * $$\begin{align}

\lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x\right)!} & = \lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x\right)!} \\ & = \lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x-1\right)!\left(w+x\right)} \\ & = \lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x-2\right)!\left(w+x-1\right)\left(w+x\right)} \\ & = \lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w\right)!\left(w+1\right)\left(w+2\right)\cdots\left(w+x-1\right)\left(w+x\right)} \\ & = \lim_{w\rightarrow\infty}\frac{\cancel{\color{OliveGreen}{w!}}\times\left(w+1\right)^x}{\cancel{\color{OliveGreen}{w!}}\left(w+1\right)\left(w+2\right)\cdots\left(w+x-1\right)\left(w+x\right)} \\ & = \lim_{w\rightarrow\infty}\frac{\left(w+1\right)^x}{\underbrace{\left(w+1\right)\left(w+2\right)\cdots\left(w+x-1\right)\left(w+x\right)}_{x}} \\ & = \lim_{w\rightarrow\infty}\frac{\overbrace{\left(w+1\right)\left(w+1\right)\cdots\left(w+1\right)\left(w+1\right)}^{x}}{\underbrace{\left(w+1\right)\left(w+2\right)\cdots\left(w+x-1\right)\left(w+x\right)}_{x}} \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{x}\frac{w+1}{w+m} \\ \lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x\right)!} & = \prod_{m=1}^{x}\lim_{w\rightarrow\infty}\frac{w+1}{w+m} \\ \end{align}$$
 * Using L'Hôpital's rule
 * $$\begin{align}

\lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x\right)!} & = \prod_{m=1}^{x}\lim_{w\rightarrow\infty}\frac{\frac{d}{dw}w+1}{\frac{d}{dw}w+m} \\ & = \prod_{m=1}^{x}\lim_{w\rightarrow\infty}\frac{1+0}{1+0} \\ & = \prod_{m=1}^{x}1 \\ \lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x\right)!} & = 1 \end{align}$$
 * Using this limit we can multiply both sides by $$x!$$:
 * $$\begin{align}

1 & = \lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\left(w+x\right)!} \\ x! & = \lim_{w\rightarrow\infty}\frac{x!\times w!\times\left(w+1\right)^x}{\left(w+x\right)!} \\ x! & = \lim_{w\rightarrow\infty}\frac{x!\times w!\times\left(w+1\right)^x}{\left(w+x-1\right)!\left(w+x\right)} \\ & = \lim_{w\rightarrow\infty}\frac{x!\times w!\times\left(w+1\right)^x}{\left(w+x-2\right)!\left(w+x-1\right)\left(w+x\right)} \\ & = \lim_{w\rightarrow\infty}\frac{x!\times w!\times\left(w+1\right)^x}{\left(x\right)!\underbrace{\left(x+1\right)\left(x+2\right)\dots\left(x+w-1\right)\left(x+w\right)}_{w}} \\ & = \lim_{w\rightarrow\infty}\frac{\cancel{\color{OliveGreen}{x!}}\times w!\times\left(w+1\right)^x}{\cancel{\color{OliveGreen}{x!}}\underbrace{\left(x+1\right)\left(x+2\right)\cdots\left(x+w-1\right)\left(x+w\right)}_{w}} \\ & = \lim_{w\rightarrow\infty}\frac{w!\times\left(w+1\right)^x}{\underbrace{\left(x+1\right)\left(x+2\right)\dots\left(x+w-1\right)\left(x+w\right)}_{w}} \\ & = \lim_{w\rightarrow\infty}\frac{\overbrace{1\times2\times3\times\cdots\times\left(w-1\right)\times\left(w\right)}^{w}}{\underbrace{\left(x+1\right)\left(x+2\right)\cdots\left(x+w-1\right)\left(x+w\right)}_{w}}\times\left(w+1\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\left(\frac{m}{x+m}\right)\;\;\times\left(w+1\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\left(\frac{m}{x+m}\right)\;\;\times\left(\left(w\right)\frac{w+1}{w}\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\left(\frac{m}{x+m}\right)\;\;\times\left(\left(w-1\right)\frac{w}{w-1}\frac{w+1}{w}\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\left(\frac{m}{x+m}\right)\;\;\times\left(\left(w-2\right)\frac{w-1}{w-2}\frac{w}{w-1}\frac{w+1}{w}\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\left(\frac{m}{x+m}\right)\;\;\times\left(\frac{2}{1}\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{w-1}{w-2}\frac{w}{w-1}\frac{w+1}{w}\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\left(\frac{m}{x+m}\right)\;\;\times\left(\underbrace{\frac{2}{1}\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{w-1}{w-2}\frac{w}{w-1}\frac{w+1}{w}}_{w}\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\left(\frac{m}{x+m}\right)\;\;\prod_{m=1}^{w}\left(\frac{m+1}{m}\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\frac{m}{x+m}\left(\frac{m+1}{m}\right)^x \\ & = \lim_{w\rightarrow\infty}\prod_{m=1}^{w}\frac{m}{x+m}\left(1+\frac{1}{m}\right)^x \\ & = \prod_{m=1}^{\infty}\frac{m}{x+m}\left(1+\frac{1}{m}\right)^x \\ x! & = \prod_{m=1}^{\infty}\frac{m}{x+m}\left(1+\frac{1}{m}\right)^x \\ \end{align}$$

Mellin transform
$$\left\{\mathcal{M}\;\frac{1}{e^x-1}\right\}\left(x\right)\;=\;\int_0^{\infty}s^{x-1}\frac{1}{e^s-1}ds \;=\; \displaystyle\Gamma\left(x\right)\displaystyle\zeta\left(x\right)$$

The Mellin transform of $$\frac{1}{e^x-1}$$
 * $$\left\{\mathcal{M}\;\frac{1}{e^x-1}\right\}\left(x\right)\;=\;\int_0^{\infty}s^{x-1}\frac{1}{e^s-1}ds$$
 * $$\begin{align}

\int_0^{\infty}s^{x-1}\frac{1}{e^s-1}ds & = \int_0^{\infty}s^{x-1}\frac{1}{e^s-1}ds \\ & = \int_0^{\infty}s^{x-1}\frac{1}{\frac{1}{e^{-s}}-1}ds \\ & = \int_0^{\infty}s^{x-1}\frac{e^{-s}}{1-e^{-s}}ds \\ & = \int_0^{\infty}s^{x-1}\sum_{n=1}^{\infty}\left(e^{-s}\right)^nds \\ & = \sum_{n=1}^{\infty}\int_0^{\infty}s^{x-1}e^{-sn}ds \\ \end{align}$$
 * $$ns=u \rightarrow s=\frac{u}{n} \rightarrow \frac{ds}{du} = \frac{1}{n} \rightarrow

\begin{cases} \text{upper bound: }\lim_{u\rightarrow\infty}\frac{u}{n}=\infty \\ \text{lower bound: }\lim_{u\rightarrow0}\frac{u}{n}=0 \end{cases}$$
 * $$\begin{align}

\int_0^{\infty}s^{x-1}\frac{1}{e^s-1}ds & = \sum_{n=1}^{\infty}\int_0^{\infty}\left(\frac{u}{n}\right)^{x-1}e^{-n\frac{u}{n}}\frac{1}{n}ds \\ & = \sum_{n=1}^{\infty}\left(\frac{1}{n}\right)^{x-1}\times\frac{1}{n}\times\int_0^{\infty}u^{x-1}e^{-u}ds \\ & = \sum_{n=1}^{\infty}\left(\frac{1}{n}\right)^x\int_0^{\infty}u^{x-1}e^{-u}ds \\ & = \sum_{n=1}^{\infty}\left(\frac{1}{n}\right)^x \displaystyle\Gamma\left(x\right) \\ & = \displaystyle\Gamma\left(x\right) \sum_{n=1}^{\infty}\left(\frac{1}{n}\right)^x \\ \int_0^{\infty}s^{x-1}\frac{1}{e^s-1}ds & = \displaystyle\Gamma\left(x\right)\displaystyle\zeta\left(x\right) \end{align}$$

Laplace Transform

 * $$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right)\;=\;\int_0^{\infty}\sin\left(t\right)e^{-xt}dt\;=\;\frac{1}{1+x^2}$$


 * $$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) = \int_0^{\infty}\sin\left(t\right)e^{-xt}dt$$

Using the Taylor series of $$\sin\left(z\right)$$ that equals $$\sum_{n=0}^{\infty}\left(-1\right)^n\frac{z^{2n+1}}{\left(2n+1\right)!}$$
 * $$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) = \int_0^{\infty}\sum_{n=0}^{\infty}\left(-1\right)^n\frac{t^{2n+1}}{\left(2n+1\right)!}e^{-xt}dt$$
 * $$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) = \sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\int_0^{\infty}t^{2n+1}e^{-xt}dt$$
 * $$xt=v\rightarrow t=\frac{v}{x}\rightarrow \frac{dt}{dv}=\frac{1}{x} \rightarrow

\begin{cases} \text{upper bound: }\lim_{v\rightarrow\infty}\frac{v}{x}=\infty \\ \text{lower bound: }\lim_{v\rightarrow0}\frac{v}{x}=0 \end{cases}$$
 * $$\begin{align}

\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) & = \sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\int_0^{\infty}\left(\frac{t}{x}\right)^{2n+1}e^{-x\frac{t}{x}}\frac{1}{x}dt \\ & = \sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\frac{1}{x}\int_0^{\infty}\left(\frac{t}{x}\right)^{2n+1}e^{-\cancel{\color{OliveGreen}{x}}\frac{t}{\cancel{\color{OliveGreen}{x}}}}dt \\ & = \sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\frac{1}{x}\left(\frac{1}{x}\right)^{2n+1}\int_0^{\infty}t^{2n+1}e^{-t}dt \\ & = \frac{1}{x^2}\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\left(\frac{1}{x^2}\right)^{n}\int_0^{\infty}t^{2n+1}e^{-t}dt \\ \end{align}$$

Using the definition $$\int_0^{\infty}t^{x}e^{-t}dt=x!$$ we get:

$$\begin{matrix} \mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\left(\frac{1}{x^2}\right)^{n}\left(2n+1\right)! \\ & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\cancel{\color{OliveGreen}{\left(2n+1\right)!}}}\left(\frac{1}{x^2}\right)^{n}\cancel{\color{OliveGreen}{\left(2n+1\right)!}} \\ & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\left(-1\right)^n\left(\frac{1}{x^2}\right)^{n} \\ & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\left(\frac{1}{-x^2}\right)^{n} \\ \end{matrix}$$

To solve this we substitute $$\frac{1}{-x^2}$$ with $$p$$

$$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}p^{n}$$

Now solving $$\displaystyle\sum_{n=0}^{\infty}p^{n}$$

$$\begin{matrix} \displaystyle\sum_{n=0}^{\infty}p^{n} & = & S \\ S & = & 1 + p + p^2 + p^3 + \cdots \\ p\times S & = & p + p^2 + p^3 + p^4 + \cdots \\ 1 + p\times S & = & 1 + p + p^2 + p^3 + p^4 + \cdots \\ 1 + p\times S & = & S \\ 1 & = & S - p\times S \\ 1 & = & S\times\left(1-p\right) \\ \displaystyle\frac{1}{1-p} & = & S \\ \displaystyle\sum_{n=0}^{\infty}p^{n} & = & \displaystyle\frac{1}{1-p} \\ \end{matrix}$$

$$\begin{matrix} \mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) & = \displaystyle\frac{1}{x^2}\frac{1}{1-p}\\ & = \displaystyle\frac{1}{x^2}\frac{1}{1-\frac{1}{-x^2}}\\ & = \displaystyle\frac{1}{x^2}\frac{x^2}{x^2+1}\\ & = \displaystyle\cancel{\color{OliveGreen}{\frac{1}{x^2}}}\frac{\cancel{\color{OliveGreen}{x^2}}}{x^2+1}\\ \mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) & = \displaystyle\frac{1}{x^2+1}\\ \end{matrix}$$

The Laplace transform of the Error function

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right)=\frac{e^{\left(\frac{x}{2}\right)^2}}{x}\left(1-\operatorname{erf}\left(\frac{x}{2}\right)\right)$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \int_0^{\infty}\operatorname{erf}\left(t\right)e^{-xt}dt $$

Integrating by parts we get:

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(\operatorname{erf}\left(t\right)\times\int e^{-xt}dt\right)_0^{\infty} - \int_{0}^{\infty}\int e^{-xt}dt\times\frac{d}{dt}\operatorname{erf}\left(t\right)dt$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(\operatorname{erf}\left(t\right)\times\int e^{-xt}dt\right)_0^{\infty} - \int_{0}^{\infty}\int e^{-xt}dt\times\frac{2}{\sqrt{\pi}}e^{-t^2}dt$$


 * $$-xt=u \rightarrow t=\frac{u}{-x} \rightarrow \frac{dt}{du}=\frac{1}{-x} \rightarrow

\begin{cases} \text{upper bound: }\lim_{u\rightarrow\infty}\frac{u}{-x}=-\infty \\ \text{lower bound: }\lim_{u\rightarrow0}\frac{u}{-x}=0 \end{cases}$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(\operatorname{erf}\left(t\right)\times\int e^{-x\frac{u}{-x}}\frac{1}{-x}dt\right)_0^{\infty} - \int_{0}^{\infty}\int e^{-x\frac{u}{-x}}\frac{1}{-x}dt\times\frac{2}{\sqrt{\pi}}e^{-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(\operatorname{erf}\left(t\right)\times\int e^{\cancel{\color{Red}{-x}}\frac{u}{\cancel{\color{Red}{-x}}}}\frac{1}{-x}du\right)_0^{\infty} - \int_{0}^{\infty}\int e^{\cancel{\color{OliveGreen}{-x}}\frac{u}{\cancel{\color{OliveGreen}{-x}}}}\frac{1}{-x}du\times\frac{2}{\sqrt{\pi}}e^{-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(-\frac{1}{x}\operatorname{erf}\left(t\right)\times\int e^{u}du\right)_0^{\infty} + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}\int e^{u}du e^{-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(-\frac{1}{x}\operatorname{erf}\left(t\right)e^{u}\right)_0^{\infty} + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{u} e^{-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(-\frac{1}{x}\operatorname{erf}\left(t\right)e^{-xt}\right)_0^{\infty} + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \lim_{w\rightarrow\infty}\left(-\frac{1}{x}\operatorname{erf}\left(w\right)e^{-x w}\right)-\lim_{h\rightarrow0}\left(-\frac{1}{x}\operatorname{erf}\left(h\right)e^{-x h}\right) + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \lim_{w\rightarrow\infty}-\left(\frac{\operatorname{erf}\left(w\right)}{xe^{x w}}\right)+\lim_{h\rightarrow0}\left(\frac{\operatorname{erf}\left(h\right)}{xe^{x h}}\right) + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

Now solving $$\lim_{w\rightarrow\infty}-\frac{\operatorname{erf}\left(w\right)}{xe^{x w}}$$:


 * The limit $$\lim_{w\rightarrow\infty}\operatorname{erf}\left(w\right) = 1$$ so


 * $$\lim_{w\rightarrow\infty}-\frac{\operatorname{erf}\left(w\right)}{xe^{x w}} = \lim_{w\rightarrow\infty}-\frac{1}{xe^{x w}}$$


 * $$\lim_{w\rightarrow\infty}-\frac{\operatorname{erf}\left(w\right)}{xe^{x w}} = 0$$

Now solving $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}}$$:


 * It can be solved by replacing $$h$$ with $$0$$:


 * $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}} = \frac{\operatorname{erf}\left(0\right)}{xe^{x\times0}}$$


 * $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}} = \frac{0}{x\times1}$$


 * $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}} = 0$$

Going back we can replace $$\lim_{w\rightarrow\infty}-\frac{\operatorname{erf}\left(w\right)}{xe^{x w}}$$ with $$0$$ and $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}}$$ with $$0$$ we get:

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \lim_{w\rightarrow\infty}-\left(\frac{\operatorname{erf}\left(w\right)}{xe^{x w}}\right)+\lim_{h\rightarrow0}\left(\frac{\operatorname{erf}\left(h\right)}{xe^{x h}}\right) + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = 0 + 0 + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(2\frac{x}{2}t+t^2\right)}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\left(\frac{x}{2}\right)^2+2\frac{x}{2}t+t^2 - \left(\frac{x}{2}\right)^2\right)}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\left(\frac{x}{2}+t\right)^2 - \left(\frac{x}{2}\right)^2\right)}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\left(\frac{x}{2}+t\right)^2\right)}e^{\left(\frac{x}{2}\right)^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\frac{x}{2}+t\right)^2}dt $$


 * $$t+\frac{x}{2}=v \rightarrow t=v-\frac{x}{2} \rightarrow \frac{dt}{dv}=1 \rightarrow

\begin{cases} \text{upper bound: }\lim_{v\rightarrow\infty}v-\frac{x}{2}=\infty \\ \text{lower bound: }\lim_{v\rightarrow0}v-\frac{x}{2}=-\frac{x}{2} \end{cases}$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\int_{-\frac{x}{2}}^{\infty}e^{-\left(\frac{x}{2}+v-\frac{x}{2}\right)^2}\times1dv $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\int_{-\frac{x}{2}}^{\infty}e^{-\left(\cancel{\color{Red}{\frac{x}{2}}}+v-\cancel{\color{Red}{\frac{x}{2}}}\right)^2}dv $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\int_{-\frac{x}{2}}^{\infty}e^{-v^2}dv $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\left(\int_{0}^{\infty}e^{-v^2}dv + \int_{-\frac{x}{2}}^{0}e^{-v^2}dv\right) $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\left(\int_{0}^{\infty}e^{-v^2}dv - \int_{0}^{-\frac{x}{2}}e^{-v^2}dv\right) $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{e^{\left(\frac{x}{2}\right)^2}}{x}\left(\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-v^2}dv - \frac{2}{\sqrt{\pi}}\int_{0}^{-\frac{x}{2}}e^{-v^2}dv\right) $$

The Error function is defined by $$\operatorname{erf}\left(k\right) = \frac{2}{\sqrt{\pi}}\int_{0}^{k}e^{-v^2}dv$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \lim_{w\rightarrow\infty}\frac{e^{\left(\frac{x}{2}\right)^2}}{x}\left(\operatorname{erf}\left(w\right) - \operatorname{erf}\left(-\frac{x}{2}\right)\right) $$

The limit $$\lim_{w\rightarrow\infty}\operatorname{erf}\left(w\right) = 1$$ and $$\operatorname{erf}\left(-k\right) = -\operatorname{erf}\left(k\right)$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{e^{\left(\frac{x}{2}\right)^2}}{x}\left(1 + \operatorname{erf}\left(\frac{x}{2}\right)\right) $$

Indefinite integrals
$$\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(A e^{Bx}\right) ^2}{2B} + \frac{ W\left(A e^{Bx}\right) }{B} + C $$

$$\int W\left(A e^{Bx}\right) \, dx \; = \; \int W\left(A e^{Bx}\right) \, dx $$
 * $$ u = Bx \rightarrow \frac{u}{B} = x \;\;\;\; \frac{ d }{ du } \frac{u}{B} = \frac{1}{B} $$

$$\int W\left(A e^{Bx}\right) \, dx \; = \; \int W\left(A e^{u}\right) \frac{1}{B} du $$
 * $$ v = e^u \rightarrow \ln\left(v\right) = u \;\;\;\; \frac{ d }{ dv } \ln\left(v\right) = \frac{1}{v} $$

$$\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int \frac{W\left(A v\right)}{v} dv $$
 * $$ w = Av \rightarrow \frac{w}{A} = v \;\;\;\; \frac{ d }{ dw } \frac{w}{A} = \frac{1}{A} $$

$$\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int \frac{\cancel{\color{OliveGreen}{A}} W(w)}{w} \cancel{\color{OliveGreen}{ \frac{1}{A} }} dw $$
 * $$ t = W\left(w\right) \rightarrow te^t=w \;\;\;\; \frac{ d }{ dt } te^t = \left(t+1\right)e^t $$

$$\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int \frac{t}{te^t}\left(t+1\right)e^t dt $$ $$\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int \frac{ \cancel{\color{OliveGreen}{t}} }{ \cancel{\color{OliveGreen}{t}} \cancel{\color{BrickRed}{e^t}} }\left(t+1\right) \cancel{\color{BrickRed}{e^t}} dt $$ $$\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int t+1 dt $$ $$ \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{t^2}{2B} + \frac{t}{B} + C $$
 * $$ t = W\left(w\right) $$

$$ \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(w\right) ^2}{2B} + \frac{ W\left(w\right) }{B} + C $$
 * $$ w = Av $$

$$ \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(Av\right) ^2}{2B} + \frac{ W\left(Av\right) }{B} + C $$
 * $$ v = e^u $$

$$ \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(Ae^u\right) ^2}{2B} + \frac{ W\left(Ae^u\right) }{B} + C $$
 * $$ u = Bx $$

$$ \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(Ae^{Bx}\right) ^2}{2B} + \frac{ W\left(Ae^{Bx}\right) }{B} + C $$

$$ \int \frac{ W(x) }{x^2} \, dx \; = \; \operatorname{Ei}\left(- W(x) \right) - e^{ - W(x) } + C $$

Introduce substitution variable $$ u = W(x)$$, which gives us $$ ue^u = x $$ and $$\frac{ d }{ du } ue^u = \left(u+1\right)e^u $$

$$\begin{align} \int \frac{ W(x) }{x^2} \, dx \; & = \; \int \frac{ u }{ \left(ue^u\right)^2 } \left(u+1\right)e^u du \\ & = \; \int \frac{ u+1 }{ ue^u } du \\ & = \; \int \frac{ u }{ ue^u } du \; + \; \int \frac{ 1 }{ ue^u } du \\ & = \; \int e^{-u} du \; + \; \int \frac{ e^{-u} }{ u } du \end{align}$$
 * $$ v = -u \rightarrow -v = u \;\;\;\; \frac{ d }{ dv } -v = -1 $$

$$ \int \frac{ W(x) }{x^2} \, dx \; = \; \int e^{v} \left(-1\right) dv \; + \; \int \frac{ e^{-u} }{ u } du $$ $$ \int \frac{ W(x) }{x^2} \, dx \; = \; - e^v + \operatorname{Ei}\left(-u\right) + C $$
 * $$ v = -u $$

$$ \int \frac{ W(x) }{x^2} \, dx \; = \; - e^{-u} + \operatorname{Ei}\left(-u\right) + C $$
 * $$ u = W(x) $$

$$\begin{align} \int \frac{ W(x) }{x^2} \, dx \; &= \; - e^{- W(x) } + \operatorname{Ei}\left(- W(x) \right) + C \\ &= \; \operatorname{Ei}\left(- W(x) \right) - e^{- W(x) } + C \end{align}$$

$$\int_0^{\infty}W_0\left(\frac{1}{t^x}\right)dt = -\left(-x\right)^{-x}\displaystyle\Gamma\left(x\right)\qquad x\isin\left\langle0;\infty\right\rangle$$

$$\int_0^{\infty}W_0\left(\frac{1}{t^x}\right)dt = \int_0^{\infty}W_0\left(\frac{1}{t^x}\right)dt$$
 * $$v=\frac{1}{t^x}\rightarrow t=v^{-\frac{1}{x}}\rightarrow\frac{dt}{dv}=-\frac{1}{x}v^{-\frac{1}{x}-1}$$

$$\begin{align} \int_0^{\infty}W_0\left(\frac{1}{t^x}\right)dt & = \int_{\infty}^{0}W\left(v\right)\left(-\frac{1}{x}v^{-\frac{1}{x}-1}\right)dv \\ & = \frac{1}{x}\int_{0}^{\infty}W\left(v\right)v^{-\frac{1}{x}-1}dv \end{align}$$


 * $$u=W\left(v\right)\rightarrow v=ue^u\rightarrow\frac{dv}{du}=\left(u+1\right)e^u$$

$$\begin{align} \int_0^{\infty}W_0\left(\frac{1}{t^x}\right)dt & = \frac{1}{x}\int_{0}^{\infty}u\left(ue^u\right)^{-\frac{1}{x}-1}\left(u+1\right)e^udu \\ & = \frac{1}{x}\int_{0}^{\infty}u^{1-\frac{1}{x}}e^{-\frac{u}{x}}+u^{-\frac{1}{x}}e^{-\frac{u}{x}}du \end{align}$$


 * $$w=\frac{u}{x}\rightarrow u=wx\rightarrow\frac{du}{dw}=x$$

$$\begin{align} \int_0^{\infty}W_0\left(\frac{1}{t^x}\right)dt & = \frac{1}{x}\int_{0}^{\infty}\left(xw\right)^{1-\frac{1}{x}}e^{-\frac{xw}{x}}+\left(xw\right)^{-\frac{1}{x}}e^{-\frac{xw}{x}}xdw \\ & = \cancel{\color{OliveGreen}{\frac{1}{x}}}\int_{0}^{\infty}\left(xw\right)^{1-\frac{1}{x}}e^{-\frac{\cancel{\color{OrangeRed}{x}}w}{\cancel{\color{OrangeRed}{x}}}}+\left(xw\right)^{-\frac{1}{x}}e^{-\frac{\cancel{\color{BlueGreen}{x}}w}{\cancel{\color{BlueGreen}{x}}}}\cancel{\color{OliveGreen}{x}}dw \\ & = \int_{0}^{\infty}x^{1-\frac{1}{x}}\times w^{1-\frac{1}{x}}e^{-w}+x^{-\frac{1}{x}}\times w^{-\frac{1}{x}}e^{-w}dw \\ & = \int_{0}^{\infty}x^{1-\frac{1}{x}}\times w^{1-\frac{1}{x}}e^{-w}dw+\int_{0}^{\infty}x^{-\frac{1}{x}}\times w^{-\frac{1}{x}}e^{-w}dw \\ & = x^{1-\frac{1}{x}}\int_{0}^{\infty}w^{1-\frac{1}{x}}e^{-w}dw+x^{-\frac{1}{x}}\int_{0}^{\infty}w^{-\frac{1}{x}}e^{-w}dw \\ \end{align}$$

$$\begin{align} p-1=1-\frac{1}{x} \rightarrow p=2-\frac{1}{x} \\ q-1=-\frac{1}{x} \rightarrow q=1-\frac{1}{x} \end{align}$$

$$\begin{align} \int_0^{\infty}W_0\left(\frac{1}{t^x}\right)dt & = x^{1-\frac{1}{x}}\int_{0}^{\infty}w^{p-1}e^{-w}dw+x^{-\frac{1}{x}}\int_{0}^{\infty}w^{q-1}e^{-w}dw \\ & = x^{1-\frac{1}{x}}\displaystyle\Gamma\left(p\right)+x^{-\frac{1}{x}}\displaystyle\Gamma\left(q\right) \\ & = x^{1-\frac{1}{x}}\displaystyle\Gamma\left(2-\frac{1}{x}\right)+x^{-\frac{1}{x}}\displaystyle\Gamma\left(1-\frac{1}{x}\right) \\ & = x^{1-\frac{1}{x}}\displaystyle\Gamma\left(1-\frac{1}{x}\right)\left(1-\frac{1}{x}\right)+x^{-\frac{1}{x}}\displaystyle\Gamma\left(1-\frac{1}{x}\right) \\ & = \displaystyle\Gamma\left(1-\frac{1}{x}\right)\left(x^{1-\frac{1}{x}}\left(1-\frac{1}{x}\right)+x^{-\frac{1}{x}}\right) \\ & = \displaystyle\Gamma\left(1-\frac{1}{x}\right)\left(x\times x^{-\frac{1}{x}}\left(1-\frac{1}{x}\right)+x^{-\frac{1}{x}}\right) \\ & = x^{-\frac{1}{x}}\displaystyle\Gamma\left(1-\frac{1}{x}\right)\left(x\left(1-\frac{1}{x}\right)+1\right) \\ & = x^{-\frac{1}{x}}\displaystyle\Gamma\left(1-\frac{1}{x}\right)\left(x-\frac{x}{x}+1\right) \\ & = x^{-\frac{1}{x}}\displaystyle\Gamma\left(1-\frac{1}{x}\right)x \\ & = x^{1-\frac{1}{x}}\displaystyle\Gamma\left(1-\frac{1}{x}\right) \\ & = x^{1-\frac{1}{x}}\displaystyle\Gamma\left(-\frac{1}{x}\right)\left(-\frac{1}{x}\right) \\ \int_0^{\infty}W_0\left(\frac{1}{t^x}\right)dt & = -x^{-\frac{1}{x}}\displaystyle\Gamma\left(-\frac{1}{x}\right)\qquad x\isin\left\langle0;\infty\right\rangle \\ \int_0^{\infty}W_0\left(t^x\right)dt & = -\left(-x\right)^{-\frac{1}{-x}}\displaystyle\Gamma\left(-\frac{1}{-x}\right)\qquad x\isin\left\langle-\infty;0\right\rangle \\ & = -\left(-x\right)^{\frac{1}{x}}\displaystyle\Gamma\left(\frac{1}{x}\right) \\ \int_0^{\infty}W_0\left(\sqrt[x]{t}\right)dt & = -\left(-x\right)^{-x}\displaystyle\Gamma\left(x\right)\qquad x\isin\left\langle-\infty;0\right\rangle \\ \end{align}$$

$$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right)=\frac{1}{1+x^2}$$

$$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) = \int_0^{\infty}\sin\left(t\right)e^{-xt}dt$$

Using the Taylor series of $$\sin\left(z\right)$$ that equals $$\sum_{n=0}^{\infty}\left(-1\right)^n\frac{z^{2n+1}}{\left(2n+1\right)!}$$

$$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) = \int_0^{\infty}\sum_{n=0}^{\infty}\left(-1\right)^n\frac{t^{2n+1}}{\left(2n+1\right)!}e^{-xt}dt$$

$$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) = \sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\int_0^{\infty}t^{2n+1}e^{-xt}dt$$


 * $$xt=v\rightarrow t=\frac{v}{x}\rightarrow \frac{dt}{dv}=\frac{1}{x}$$

$$\begin{matrix} \mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) & = \displaystyle\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\int_0^{\infty}\left(\frac{t}{x}\right)^{2n+1}e^{-x\frac{t}{x}}\frac{1}{x}dt \\ & = \displaystyle\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\frac{1}{x}\int_0^{\infty}\left(\frac{t}{x}\right)^{2n+1}e^{-\cancel{\color{OliveGreen}{x}}\frac{t}{\cancel{\color{OliveGreen}{x}}}}dt \\ & = \displaystyle\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\frac{1}{x}\left(\frac{1}{x}\right)^{2n+1}\int_0^{\infty}t^{2n+1}e^{-t}dt \\ & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\left(\frac{1}{x^2}\right)^{n}\int_0^{\infty}t^{2n+1}e^{-t}dt \\ \end{matrix}$$

Using the definition $$\int_0^{\infty}t^{x}e^{-t}dt=x!$$ we get:

$$\begin{matrix} \mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)!}\left(\frac{1}{x^2}\right)^{n}\left(2n+1\right)! \\ & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\cancel{\color{OliveGreen}{\left(2n+1\right)!}}}\left(\frac{1}{x^2}\right)^{n}\cancel{\color{OliveGreen}{\left(2n+1\right)!}} \\ & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\left(-1\right)^n\left(\frac{1}{x^2}\right)^{n} \\ & = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}\left(\frac{1}{-x^2}\right)^{n} \\ \end{matrix}$$

To solve this we substitute $$\frac{1}{-x^2}$$ with $$p$$

$$\mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) = \displaystyle\frac{1}{x^2}\sum_{n=0}^{\infty}p^{n}$$

Now solving $$\displaystyle\sum_{n=0}^{\infty}p^{n}$$

$$\begin{matrix} \displaystyle\sum_{n=0}^{\infty}p^{n} & = & S \\ S & = & 1 + p + p^2 + p^3 + \cdots \\ p\times S & = & p + p^2 + p^3 + p^4 + \cdots \\ 1 + p\times S & = & 1 + p + p^2 + p^3 + p^4 + \cdots \\ 1 + p\times S & = & S \\ 1 & = & S - p\times S \\ 1 & = & S\times\left(1-p\right) \\ \displaystyle\frac{1}{1-p} & = & S \\ \displaystyle\sum_{n=0}^{\infty}p^{n} & = & \displaystyle\frac{1}{1-p} \\ \end{matrix}$$

$$\begin{matrix} \mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) & = \displaystyle\frac{1}{x^2}\frac{1}{1-p}\\ & = \displaystyle\frac{1}{x^2}\frac{1}{1-\frac{1}{-x^2}}\\ & = \displaystyle\frac{1}{x^2}\frac{x^2}{x^2+1}\\ & = \displaystyle\cancel{\color{OliveGreen}{\frac{1}{x^2}}}\frac{\cancel{\color{OliveGreen}{x^2}}}{x^2+1}\\ \mathcal{L}\left\{\sin\left(t\right)\right\}\left(x\right) & = \displaystyle\frac{1}{x^2+1}\\ \end{matrix}$$

The Laplace transform of the Error function

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right)=\frac{e^{\left(\frac{x}{2}\right)^2}}{x}\left(1-\operatorname{erf}\left(\frac{x}{2}\right)\right)$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \int_0^{\infty}\operatorname{erf}\left(t\right)e^{-xt}dt $$

Integrating by parts we get:

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(\operatorname{erf}\left(t\right)\times\int e^{-xt}dt\right)_0^{\infty} - \int_{0}^{\infty}\int e^{-xt}dt\times\frac{d}{dt}\operatorname{erf}\left(t\right)dt$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(\operatorname{erf}\left(t\right)\times\int e^{-xt}dt\right)_0^{\infty} - \int_{0}^{\infty}\int e^{-xt}dt\times\frac{2}{\sqrt{\pi}}e^{-t^2}dt$$


 * $$-xt=u \rightarrow t=\frac{u}{-x} \rightarrow \frac{dt}{du}=\frac{1}{-x} \rightarrow

\begin{cases} \text{upper bound: }\lim_{u\rightarrow\infty}\frac{u}{-x}=-\infty \\ \text{lower bound: }\lim_{u\rightarrow0}\frac{u}{-x}=0 \end{cases}$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(\operatorname{erf}\left(t\right)\times\int e^{-x\frac{u}{-x}}\frac{1}{-x}dt\right)_0^{\infty} - \int_{0}^{\infty}\int e^{-x\frac{u}{-x}}\frac{1}{-x}dt\times\frac{2}{\sqrt{\pi}}e^{-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(\operatorname{erf}\left(t\right)\times\int e^{\cancel{\color{Red}{-x}}\frac{u}{\cancel{\color{Red}{-x}}}}\frac{1}{-x}du\right)_0^{\infty} - \int_{0}^{\infty}\int e^{\cancel{\color{OliveGreen}{-x}}\frac{u}{\cancel{\color{OliveGreen}{-x}}}}\frac{1}{-x}du\times\frac{2}{\sqrt{\pi}}e^{-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(-\frac{1}{x}\operatorname{erf}\left(t\right)\times\int e^{u}du\right)_0^{\infty} + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}\int e^{u}du e^{-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(-\frac{1}{x}\operatorname{erf}\left(t\right)e^{u}\right)_0^{\infty} + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{u} e^{-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \left(-\frac{1}{x}\operatorname{erf}\left(t\right)e^{-xt}\right)_0^{\infty} + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \lim_{w\rightarrow\infty}\left(-\frac{1}{x}\operatorname{erf}\left(w\right)e^{-x w}\right)-\lim_{h\rightarrow0}\left(-\frac{1}{x}\operatorname{erf}\left(h\right)e^{-x h}\right) + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \lim_{w\rightarrow\infty}-\left(\frac{\operatorname{erf}\left(w\right)}{xe^{x w}}\right)+\lim_{h\rightarrow0}\left(\frac{\operatorname{erf}\left(h\right)}{xe^{x h}}\right) + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

Now solving $$\lim_{w\rightarrow\infty}-\frac{\operatorname{erf}\left(w\right)}{xe^{x w}}$$:


 * The limit $$\lim_{w\rightarrow\infty}\operatorname{erf}\left(w\right) = 1$$ so


 * $$\lim_{w\rightarrow\infty}-\frac{\operatorname{erf}\left(w\right)}{xe^{x w}} = \lim_{w\rightarrow\infty}-\frac{1}{xe^{x w}}$$


 * $$\lim_{w\rightarrow\infty}-\frac{\operatorname{erf}\left(w\right)}{xe^{x w}} = 0$$

Now solving $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}}$$:


 * It can be solved by replacing $$h$$ with $$0$$:


 * $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}} = \frac{\operatorname{erf}\left(0\right)}{xe^{x\times0}}$$


 * $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}} = \frac{0}{x\times1}$$


 * $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}} = 0$$

Going back we can replace $$\lim_{w\rightarrow\infty}-\frac{\operatorname{erf}\left(w\right)}{xe^{x w}}$$ with $$0$$ and $$\lim_{h\rightarrow0}\frac{\operatorname{erf}\left(h\right)}{xe^{x h}}$$ with $$0$$ we get:

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \lim_{w\rightarrow\infty}-\left(\frac{\operatorname{erf}\left(w\right)}{xe^{x w}}\right)+\lim_{h\rightarrow0}\left(\frac{\operatorname{erf}\left(h\right)}{xe^{x h}}\right) + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = 0 + 0 + \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-xt-t^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(2\frac{x}{2}t+t^2\right)}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\left(\frac{x}{2}\right)^2+2\frac{x}{2}t+t^2 - \left(\frac{x}{2}\right)^2\right)}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\left(\frac{x}{2}+t\right)^2 - \left(\frac{x}{2}\right)^2\right)}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\left(\frac{x}{2}+t\right)^2\right)}e^{\left(\frac{x}{2}\right)^2}dt $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\frac{x}{2}+t\right)^2}dt $$


 * $$t+\frac{x}{2}=v \rightarrow t=v-\frac{x}{2} \rightarrow \frac{dt}{dv}=1 \rightarrow

\begin{cases} \text{upper bound: }\lim_{v\rightarrow\infty}v-\frac{x}{2}=\infty \\ \text{lower bound: }\lim_{v\rightarrow0}v-\frac{x}{2}=-\frac{x}{2} \end{cases}$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\int_{-\frac{x}{2}}^{\infty}e^{-\left(\frac{x}{2}+v-\frac{x}{2}\right)^2}\times1dv $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\int_{-\frac{x}{2}}^{\infty}e^{-\left(\cancel{\color{Red}{\frac{x}{2}}}+v-\cancel{\color{Red}{\frac{x}{2}}}\right)^2}dv $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\int_{-\frac{x}{2}}^{\infty}e^{-v^2}dv $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\left(\int_{0}^{\infty}e^{-v^2}dv + \int_{-\frac{x}{2}}^{0}e^{-v^2}dv\right) $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{2e^{\left(\frac{x}{2}\right)^2}}{x\sqrt{\pi}}\left(\int_{0}^{\infty}e^{-v^2}dv - \int_{0}^{-\frac{x}{2}}e^{-v^2}dv\right) $$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{e^{\left(\frac{x}{2}\right)^2}}{x}\left(\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-v^2}dv - \frac{2}{\sqrt{\pi}}\int_{0}^{-\frac{x}{2}}e^{-v^2}dv\right) $$

The Error function is defined by $$\operatorname{erf}\left(k\right) = \frac{2}{\sqrt{\pi}}\int_{0}^{k}e^{-v^2}dv$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \lim_{w\rightarrow\infty}\frac{e^{\left(\frac{x}{2}\right)^2}}{x}\left(\operatorname{erf}\left(w\right) - \operatorname{erf}\left(-\frac{x}{2}\right)\right) $$

The limit $$\lim_{w\rightarrow\infty}\operatorname{erf}\left(w\right) = 1$$ and $$\operatorname{erf}\left(-k\right) = -\operatorname{erf}\left(k\right)$$

$$\mathcal{L}\left\{\operatorname{erf}\left(t\right)\right\}\left(x\right) = \frac{e^{\left(\frac{x}{2}\right)^2}}{x}\left(1 + \operatorname{erf}\left(\frac{x}{2}\right)\right) $$