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{\bf Exam Question

Topic: GammaFunction}

Using appropriate substitutions, show that

\begin{eqnarray*}
\mathrm{(i)}\ && \ \int_0^1 \ln\left(\frac{1}{u}\right)^{x-1}\,
du=\Gamma(x),\\ \mathrm{(ii)}\ && \
\int_0^{\pi/2}\left(\tan^9x+\tan^{11}x\right)\exp\left(-\tan^2x\right)\,
dx=12
\end{eqnarray*}
\vspace{0.5in}

{\bf Solution}

(i) Let $u=\mathrm{e}^{-t},$ so
$\ln\displaystyle{\ln\left(\frac{1}{u}\right)=t.}$ $$\int_0^1
\ln\left(\frac{1}{u}\right)^{x-1}\, du=\int_{\infty}^0
t^{x-1}(-\mathrm{e}^{-t})\, dt =\int_0^{\infty}
t^{x-1}(-\mathrm{e}^{-t})\, dt = \Gamma(x).$$

\begin{eqnarray*}
\mathrm{(ii)}\ \
I&=&\int_0^{\pi/2}\left(\tan^9x+\tan^{11}x\right)\exp\left(-\tan^2x\right)\,
dx\\&=&\frac{1}{2}\int_0^{\pi/2}\left(\tan^2x\right)^4.2\tan
x(1+\tan^2x) \exp\left(-\tan^2x\right)\,
dx\\&=&\frac{1}{2}\int_0^{\pi/2}\left(\tan^2x\right)^4.2\tan
x\sec^2x \exp\left(-\tan^2x\right)\, dx.
\end{eqnarray*}

Let $t=\tan^2x;\ \ dt=2\tan x\sec^2 x\, dx$ $$\mathrm{So}\ \
I=\frac{1}{2}\int_0^{\infty}t^4\mathrm{e}^{-t}\,
dt=\frac{\Gamma(5)}{2}=\frac{4!}{2}=12.$$



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