User:Cybersnoopy/test

test
$$ \begin{align} \Gamma(a,x) &= \int_x^\infty t^{a-1} \exp(-t) dt \\ Q(a,x) &= 1/\Gamma(a) \int_x^\infty t^{a-1} \exp(-t) dt \end{align} $$

sample
Change (28) to Indefinite Integral

$$ \begin{align} y & = A(t)\exp \int \frac{\partial x}{x+B(t)\exp[\frac{1}{\bar{\sigma}}\int \partial \log x]}\\ & = A(t)\exp \int \frac{x^{-1}\partial x}{1+B(t)x^{-1}\{\exp[\int \partial \log x]\}^\frac{1}{\bar{\sigma}}}\\ & = A(t)\exp \int \frac{x^{-1}\partial x}{1+B(t)x^{-1}\{\exp(\log x)\}^\frac{1}{\bar{\sigma}}}\\ & = A(t)\exp \int \frac{x^{-1}dx}{1+B(t)x^{\frac{1}{\bar{\sigma}}-1}}\\ & = A(t)\exp \int \frac{d(\log x)}{1+B(t)\exp[(\frac{1}{\bar{\sigma}}-1)\log x]} \end{align} $$

Let $$w=\log x$$, then the formular turns to be

$$y = A(t)\exp \int \frac{dw}{1+B(t)\exp[(\frac{1}{\bar{\sigma}}-1)w]}$$

On the other hand, we have (adapted from the Integral Formula Table)

$$\int \frac{dx}{a+be^{mx}}=\tfrac{1}{am}[mx-\log(a+be^{mx})]$$

In this case, $$a=1$$,$$b=B(t)$$ and $$m=\frac{1}{\bar{\sigma}}-1$$, use the formula above, we get

$$ \begin{align} y(x,t) & = A(t) \exp\{ \frac{1}{\frac{1}{\bar{\sigma}}-1} [(\frac{1}{\bar{\sigma}}-1)w - \log(1+B(t)e^{\frac{1}{(\bar{\sigma}}-1)w})]\}\\ & = A(t) \exp\{ \frac{1}{\frac{1}{\bar{\sigma}}-1} [(\frac{1}{\bar{\sigma}}-1)\log x - \log(1+B(t)e^{\frac{1}{(\bar{\sigma}}-1)\log x})]\}\\ & = A(t) \exp\{ \frac{-1}{\frac{1}{\bar{\sigma}}-1} [-(\frac{1}{\bar{\sigma}}-1)\log x + \log(1+B(t)x^{(\frac{1}{\bar{\sigma}}-1)})]\}\\ & = A(t) [\exp\{ -\log x^{\frac{1}{\bar{\sigma}}-1} + \log(1+B(t)x^{\frac{1}{(\bar{\sigma}}-1)})\}]^{\frac{-1}{\frac{1}{\bar{\sigma}}-1}}\\ & = A(t) [\exp\{\log(\frac{1+B(t)x^{\frac{1}{(\bar{\sigma}}-1)}}{x^{\frac{1}{\bar{\sigma}}-1}})\}]^{\frac{-\bar{\sigma}}{1-\bar{\sigma}}}\\ & = A(t) (x^{-(\frac{1-\bar{\sigma}}{\bar{\sigma}})}+B(t))^{-(\frac{\bar{\sigma}}{1-\bar{\sigma}})}\\ \end{align} $$