Summer 2017 – Q.1 [Preview]

1. a) Prove that $latex \int_2^1x^{n-1}\left(\log\frac12\right)^{m-1}dx=\frac{\left|\overline m\right.}{n^m}$ [6M]

Consider,
$latex \int_{0}^{1}x^{n-1}\left ( log\frac{1}{x} \right )^{m-1}dx$

put,

$latex log\frac{1}{x}=z$

$latex \therefore \frac{1}{x}=e^{z}$

$latex x=e^{-z}$

$latex dx=-e^{-z}dz$

when x→0, z→∞

when x→1, z→0

$latex =\int_{\infty }^{0}\left ( e^{-z} \right )^{n-1}z^{m-1}\left ( -e^{-z} \right )dz$

$latex =\int_{0}^{\infty}\left ( e^{-z} \right )^{n-1+1}z^{m-1}dz$

$latex =\int_{0}^{\infty}e^{-nz}z^{m-1}dz$

put nz=u

$latex z=\frac{u}{n}$;

$latex dz=\frac{du}{n}$

when z→0, u→0

when z→∞, u→∞

$latex =\int_{0}^{\infty}e^{-u}\left ( \frac{u}{n} \right )^{m-1}\frac{du}{n}$

$latex =\int_{0}^{\infty}e^{-u}u^{m-1}\frac{1}{n^{m-1+1}}du$

$latex =\frac{1}{n^{m}}\int_{0}^{\infty }e^{-u}u^{m-1}du$

$latex \therefore\left ( \int_{0}^{\infty }e^{-x}x^{m-1}dx=\sqrt{m}\right )$

$latex =\frac{1}{n^{m}}\sqrt{m}$

$latex \therefore \int_{0}^{1}x^{n-1}\left ( log\frac{1}{x} \right )^{m-1}dx=\frac{\sqrt{m}}{n^{m}}$

b) Evaluate $latex \int_0^1\frac{x^\alpha-1}{\log x}dx,\;\alpha\geq0$ by differentiating under integral sign. [6M]

Let $latex f\left ( \alpha \right )=\int_{0}^{1}\frac{x^{\alpha }-1}{logx}dx —–\left ( i \right )$

By using leibnitz’s rule

$latex f’\left ( \alpha \right )=\frac{d}{d\alpha }f\left ( \alpha \right )$

$latex =\int_{0}^{1}\frac{\partial }{\partial \alpha }\left [\frac{x^{\alpha }-1}{logx} \right ]dx$

$latex \left (\therefore \frac{d}{dx}a^{x}-1=a^{x}logx \right )$

$latex =\int_{0}^{1}\frac{x^{\alpha }logx}{logx}dx$

$latex \left (\therefore \int x^{n}=\frac{x^{n+1}}{n+1} \right )$

$latex =\int_{0}^{1}x^{\alpha }dx$

$latex =\left | \frac{\alpha ^{\alpha +1}}{\alpha +1} \right |_{o}^{1}$

$latex =\frac{1^{\alpha +1}}{\alpha +1}-\frac{0^{\alpha +1}}{\alpha +1}$

$latex =\frac{1^{\alpha +1}}{\alpha +1}-0$

$latex \left ( \therefore 1^{n} =1\right )$

$latex =\frac{1}{\alpha +1}$

Now integrating both side w.r. to α

$latex f\left ( \alpha \right )=log\left ( 1+\alpha \right )+c —–\left ( ii \right )$

$latex \left (\because \int \frac{1}{x}=logx+c \right )$

from (i) when α=0

$latex f\left ( 0 \right )=\int_{0}^{1}\frac{x^{0}-1}{logx}dx$

$latex \int_{0}^{1}\frac{1-1}{logx}dx$

$latex f\left ( 0 \right )=0$

from (ii) $latex f\left ( 0 \right )=log(1)+c$
i.e. C=0

Put value of C in equation (ii)
Hence equation (ii) gives

$latex f\left ( \alpha \right )=log(1+\alpha)$