$\int^1_0 \frac{x^a – 1}{log x} dx$

Solution:

let $\quad I=\int_{0}^{1} \frac{x^{a}-1}{\log x} d x$

Taking 'a' as parameter,

$\quad I(a)=\int_{0}^{1} \frac{x^{a}-1}{\log x} d x---(1)$

differentiate w.r.t a

$\frac{d I(a)}{d a}=\frac{d}{d a} \int_{0}^{1} \frac{x^{a}-1}{\log x} d x$

$\therefore \frac{d I(a)}{d a}=\int_{0}^{1} \frac{\partial}{\partial a} \frac{x^{a}-1}{\log x} d x \quad \ldots \ldots\{$ D.U.l.S $f(x)\}$

$\therefore \frac{d I(a)}{d a}=\int_{0}^{1} \frac{x^{a} \log x}{\log x} \mathrm{d} x \quad \ldots \ldots .\left\{\frac{d x^{a}}{d a}=x^{a} . \log a\right\}$

$\therefore \quad \frac{d I(a)}{d a}=\int_{0}^{1} x^{a} d x$

$\begin{aligned} \therefore \frac{d I(a)}{d a} &=\left[\frac{x^{a+1}}{a+1}\right]_{0}^{1} \\ \therefore \frac{d I(a)}{d a} &=\frac{1}{a+1}-0 \\ \therefore \frac{d I(a)}{d a} &=\frac{1}{a+1} \end{aligned}$

$\begin{aligned} \text { now, integrate } & \text { w.r.t } a \\ I(a) &=\int \frac{1}{a+1} d a \\ & I(a)=\log (a+1)+c ---(2)\end{aligned}$

where c is constant of integration

put $a=0$ in eqn $(1)$

$\quad I(0)=\int_{0}^{1} 0 d x=0$

And

From eqn $(2), \quad I(0)=c$

$\therefore c=0$

$\therefore I=\log (a+1)$