User:Harry Princeton/Chain Integrals and Higher-Order Integrals

This is my work on Chain Integration and Integration by Parts (higher-order integrals). Also included is a differential analysis on cycloids.

Chain Integration Formula and Examples
See Fresnel integral.

We could approximate the tangent and secant integrals

$$T(x):=\int_0^x\tan\big(t^2\big)~\text{d}t$$ and $$\text{Sc}(x):=\int_0^x\sec\big(t^2\big)~\text{d}t$$

by using the Cauchy Principal Value and integration by parts:

$$T(x)=\frac{\ln|\sec x^2|}{2x}+\int_0^x\frac{\ln|\sec t^2|~\text{d}t}{2t^2}$$ and $$\text{Sc}(x)=\frac{\ln|\tan x^2+\sec x^2|}{2x}+\int_0^x\frac{\ln|\tan t^2+\sec t^2|~\text{d}t}{2t^2}$$

with the square Chain Integration formula $$\int_0^xf\big(t^2\big)~\text{d}t=\frac{F\big(x^2\big)}{2x}+\frac{\int_0^x F\big(t^2\big)~\text{d}t}{2x^2}$$ where $$F(x)=\int_0^xf(t)~\text{d}t$$. {| class="wikitable mw-collapsible mw-collapsed" role="presentation"
 * General Chain Integration Formula For Strictly Monotone Functions $$g~$$
 * $$$$\Longleftrightarrow\frac{\text{d}}{\text{d}x}\left(\int_0^xf\big(g(t)\big)~\text{d}t\right)=\frac{\text{d}}{\text{d}x}\left(\frac{F\big(g(x)\big)}{g'(x)}+\int_0^x\frac{ F\big(g(t)\big)g(t)~\text{d}t}{g'(t)^2}\right)=\frac{F'\big(g(x)\big)g'(x)^2-F\big(g(x)\big)g(x)}{g'(x)^2}+\frac{ F\big(g(x)\big)g''(x)}{g'(x)^2}=F'\big(g(x)\big)=f\big(g(x)\big)$$as expected, using the Chain Rule, antiderivative, quotient rule, and cancellation.
 * $$$$\Longleftrightarrow\frac{\text{d}}{\text{d}x}\left(\int_0^xf\big(g(t)\big)~\text{d}t\right)=\frac{\text{d}}{\text{d}x}\left(\frac{F\big(g(x)\big)}{g'(x)}+\int_0^x\frac{ F\big(g(t)\big)g(t)~\text{d}t}{g'(t)^2}\right)=\frac{F'\big(g(x)\big)g'(x)^2-F\big(g(x)\big)g(x)}{g'(x)^2}+\frac{ F\big(g(x)\big)g''(x)}{g'(x)^2}=F'\big(g(x)\big)=f\big(g(x)\big)$$as expected, using the Chain Rule, antiderivative, quotient rule, and cancellation.

Higher-Order Integrals:
$$⇭⇭⇭These can be proved using induction by integration by parts. In fact, integrals through CPV are defined when subtracting balanced pole functions - pure powers of $$1/x$$ - from original poles, results in functions with remaining possible singularities of size $$O(n^{-\alpha})$$ with $$\alpha<1$$ strict).
 * }

Notice that both $$\tan x^2$$ and $$\sec x^2$$ have simple nonzero isolated poles, limiting to scalar multiples of $$[\mp 1/x]_{x=0}$$. Then $$\int_1^x\mp\frac{\text{d}t}{t}=\mp\ln x$$ and $$\int_0^x\mp\ln t~\text{d}t=\mp x(\ln x-1)$$with $$\lim_{x\rightarrow 0}\mp\big[x(\ln x-1)\big]=0$$.

Therefore,  $$\frac{\int_0^x F\big(t^2\big)~\text{d}t}{2x^2}$$ is bounded for $$F(x)=\ln|\sec x|=\int_0^x\tan t~\text{d}t$$ and $$F(x)=\ln|\tan x+\sec x|=\int_0^x\sec t~\text{d}t$$; as we have the related bounded integral$$\int_0^1\mp\ln x~\text{d}x=\mp[x(\ln x-1)]_0^1=\pm 1$$ and we can do scalar multiple comparisons. So by CPV, the above integrals are defined except at isolated poles. Graphs of these integrals for $$x\ge 0$$ are found below: Finally, we have, approximately:

$$\liminf_{x\rightarrow\infty}\int_0^x\tan\big(t^2\big)~\text{d}t=\lim_{x\rightarrow\infty}\int_0^x\frac{\ln|\sec t^2|~\text{d}t}{2t^2}\approx 0.535914345467$$

$$\underset{x\rightarrow\infty}{\text{lim inflection}}\int_0^x\sec\big(t^2\big)~\text{d}t=\lim_{x\rightarrow\infty}\int_0^x\frac{\ln|\tan t^2+\sec t^2|~\text{d}t}{2t^2}\approx 0.836834795146$$

compared to $$\sqrt{\pi/8}\approx 0.62665706866$$ for both $$S(x)$$ and $$C(x)$$.

Modulated Integrals
We could also approximate the cotangent and cosecant integrals

$$\text{Ct}(x):=\int_0^x\cot\big(t^2\big)~\text{d}t$$ and $$\text{Cc}(x):=\int_0^x\csc\big(t^2\big)~\text{d}t$$

by using an integration by parts, but need to isolate a pole of order 2 at $$x=0$$ for each function. We do so by subtracting $$1/x^2$$ from each function to yield bounded functions $$f(x)$$ at $$x=0$$ (in fact, with $$\lim_{x\rightarrow 0}f(x)=0$$ for both functions!), applying the same treatment $$\int_0^xf\big(t^2\big)~\text{d}t=\frac{F\big(x^2\big)}{2x}+\frac{\int_0^x F\big(t^2\big)~\text{d}t}{2x^2}$$ as previously (it works similarly at all other poles), and then adding the antiderivative $$-1/x$$ of $$1/x^2$$ back.

Since both integrals have right limit $$-\infty$$ as $$x$$ approaches $$0$$, we instead add constants so that critical points/inflection points of the cotangent/cosecant integrals, respectively, approach $$y=0$$ as $$x\rightarrow\infty$$ (canonicalization). We thus have:

$$\text{Ct}(x):\approx 1.29414-\left(\frac{\ln2-\ln\left(\left|\cot x^2+\csc x^2\right|\right)-2\ln\left(\left|x\right|\right)}{2x^2}+\frac{1}{x}+\int_0^x\frac{\ln2-\ln\left(\left|\cot t^2+\csc t^2\right|\right)-2\ln\left(\left|t\right|\right)}{2t^2}dt\right)$$

$$\text{Cc}(x):\approx \frac{\ln2-\ln\left(\left|\cot x^2+\csc x^2\right|\right)-2\ln\left(\left|x\right|\right)}{2x}-\frac{1}{x}+\int_0^x\frac{\ln2-\ln\left(\left|\cot t^2+\csc t^2\right|\right)-2\ln\left(\left|t\right|\right)}{2t^2}dt+0.5360919$$

With graphs seen below:

Higher Order (Stacked) Integrals
A general formula for the second-order real integral (second antiderivative) is

$$\iint_0^x f(t)~\text{d}t^2=x\int_0^xf(t)~\text{d}t-\int_0^xtf(t)~\text{d}t$$ A general formula for the third-order real integral (third antiderivative) is

$$\iiint_0^x f(t)~\text{d}t^3=\frac{1}{2}\left(x^2\int_0^xf(t)~\text{d}t-2x\int_0^xtf(t)~\text{d}t+\int_0^xt^2f(t)~\text{d}t\right)$$ A general formula for the $$n$$th-order real integral ($$n$$th antiderivative) is

$$\underset{n~\text{times}}{\int\cdots\int_0^x} f(t)~\text{d}t^n=\frac{1}{(n-1)!}\sum_{i=1}^n(-1)^i\binom{n-1}{i}\left[x^{n-1-i}\int_0^xt^if(t)~\text{d}t\right]$$ So to compute higher-order integrals, no other nonelementary integrals need to be considered except for those possibly equal to $$\int_0^xt^nf(t)~\text{d}t$$ for $$f\in\mathcal{R},~n\in\mathbb{N}$$ (Riemann-integrable functions). To compute from another bound $$c\ne 0$$, it is sufficient for integrals only of the form $$\int_0^x(t-c)^nf(t)~\text{d}t$$ to be considered.

Example
Below, the first five antiderivatives of $$f(x)=e^{x^2+1/x^2}$$ are computed and graphed, using this method. Since $$f$$ is undefined at $$x=0$$ (in fact, $$x=0$$ is an essential singularity, and every antiderivative diverges as $$x\rightarrow 0$$), we start at $$x=1$$ instead (a critical minimum point for $$f$$, so most suitable):


 * $$e^{x^2+1/x^2}$$
 * $$\int_1^xe^{t^2+1/t^2}~\text{d}t$$
 * $$\iint_1^xe^{t^2+1/t^2}~\text{d}t^2=(x-1)\int_1^xe^{t^2+1/t^2}~\text{d}t-\int_1^x(t-1)e^{t^2+1/t^2}~\text{d}t$$
 * $$\iiint_1^xe^{t^2+1/t^2}~\text{d}t^3=\frac{1}{2}\left[(x-1)^2\int_1^xe^{t^2+1/t^2}~\text{d}t-2(x-1)\int_1^x(t-1)e^{t^2+1/t^2}~\text{d}t+\int_1^x(t-1)^2e^{t^2+1/t^2}~\text{d}t\right]$$
 * $$\underset{4~\text{times}}{\int\cdots\int_1^x}e^{t^2+1/t^2}~\text{d}t^4=\frac{1}{6}\left[(x-1)^3\int_1^xe^{t^2+1/t^2}~\text{d}t-3(x-1)^2\int_1^x(t-1)e^{t^2+1/t^2}~\text{d}t+3(x-1)\int_1^x(t-1)^2e^{t^2+1/t^2}~\text{d}t-\int_1^x(t-1)^3e^{t^2+1/t^2}~\text{d}t\right]$$
 * $$\underset{5~\text{times}}{\int\cdots\int_1^x}e^{t^2+1/t^2}~\text{d}t^5=\frac{1}{24}\left[(x-1)^4\int_1^xe^{t^2+1/t^2}~\text{d}t-4(x-1)^3\int_1^x(t-1)e^{t^2+1/t^2}~\text{d}t+6(x-1)^2\int_1^x(t-1)^2e^{t^2+1/t^2}~\text{d}t-4(x-1)\int_1^x(t-1)^3e^{t^2+1/t^2}~\text{d}t+\int_1^x(t-1)^4e^{t^2+1/t^2}~\text{d}t\right]$$

Note that the general chain integral formula cannot be used since (1) $$g(x)=x^2+1/x^2$$ is not strictly monotone, (2) no balanced essential singularity can be isolated from $$e^{x^2+1/x^2}$$, not even $$e^{1/x^2}$$, and (3) even the balanced essential singularity $$e^{1/x^2}$$ is not elementary-integrable.

Cycloid
The work-energy formula is intrinsic to any twice continuously differentiable increasing function:

$$\frac{y'(x|_{y=b})^2-y'(x|_{y=a})^2}{2}=\int_{y=a}^{y=b}y''(x|_y)~\text{d}y$$

This is by Chain Rule, since $$\frac{\text{d}f'(f^{-1}(x))^2}{2~\text{d}x}=f''(f^{-1}(x))$$ for any twice continuously differentiable increasing function $$f$$.

Using the work-energy formula

$$\text{KE}=\frac{1}{2}m(v_f^2-v_0^2)=\int_{x_1}^{x_2}F(x)~\text{d}x$$

with $$m=1$$, initial position $$x_1=r$$, and initial velocity $$v_0(x_1)=0$$; we have

$$\left(\frac{\text{d}y}{\text{d}x}\right)^2=\frac{2r}{y}-1=v^2=2\int_r^yF(x)~\text{d}x\Longrightarrow\frac{\text{d}}{\text{d}y}\left[\frac{2r}{y}-1\right]=\frac{\text{d}}{\text{d}y}\left[2\int_r^yF(x)~\text{d}x\right]\Longleftrightarrow-\frac{2r}{y^2}=2F(y)\Longrightarrow F(y)=-\frac{r}{y^2}$$

by the Fundamental Theorem of Calculus I. So in fact the cycloid is the solution $$\big(x,y(x)\big)$$ to Newton's Gravitational Law (where time is measured in $$x$$) if a particle is bounded in a heavy point object's gravitational field (negative net energy), with zero angular momentum; if we set $$r=GM$$, $$y=R$$, to obtain $$F(R)=-\frac{GMm}{R^2}\Longrightarrow a(R)=-\frac{GM}{R^2}$$.

Other zero-angular-momentum solutions for identical mass include $$y(x)=x^{\frac{2}{3}}\sqrt[3]{\frac{9r}{2}}$$ for zero net energy, and $$\big(x,y(x)\big)=\big(r(\sinh t-t),r(\cosh t-1)\big)$$ for positive net energy. For the first case, indeed

$$\left(\frac{\text{d}y}{\text{d}x}\right)^2=\left[\frac{\text{d}}{\text{d}x}\left(x^{\frac{2}{3}}\sqrt[3]{\frac{9r}{2}}\right)\right]^2=\left[\frac{2}{3}\sqrt[3]{\frac{9r}{2x}}\right]^2=\frac{4}{9}\sqrt[3]{\frac{81r^2}{4x^2}}=\frac{2r}{\sqrt[3]{\frac{9^3}{81}\cdot\frac{4}{2^3}\cdot r}}=\frac{2r}{\sqrt[3]{\frac{9r}{2}}x^{\frac{2}{3}}}=\frac{2r}{y}$$

by the Power Rule, where the additional constant of $$-1$$ is specific to the cycloid only (which is the total energy); and for the second case, indeed

$$\left(\frac{\text{d}y}{\text{d}x}\right)^2=\left[\frac{\frac{\text{d}}{\text{d}t}\big(r(\cosh t-1)\big)}{\frac{\text{d}}{\text{d}t}\big(r(\sinh t-t)\big)}\right]^2=\left[\frac{r\sinh t}{r(\cosh t-1)}\right]^2=\left[\frac{\sinh t}{\cosh t-1}\right]^2=\frac{\sinh^2t}{\cosh^2t-2\cosh t+1}=\frac{\cosh^2t-1}{\cosh^2t-2\cosh t+1}$$

$$=\frac{(\cosh t-1)(\cosh t+1)}{(\cosh t-1)^2}=\frac{\cosh t+1}{\cosh t-1}=\frac{2}{\cosh t-1}+1=\frac{2r}{y}+1$$

by the parametric derivative, and the additional constant of $$+1$$ is specific to this second path only (which is the total energy).