October 7th, 2019, 02:05 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  Cannot solve integral
Evaluate: $\displaystyle \int_{0}^{1} \frac{x^{b}x^{a}}{\ln(x)}dx\;$ . $\displaystyle a\neq b$, $\displaystyle a,b>0.$ 
October 7th, 2019, 04:21 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  The integral is similar to $\displaystyle I(z) =\int_{0}^{1} \frac{x^z }{\ln x} dx=\frac{1}{z+1} \int_{\infty}^{0} \frac{e^{t(z+1)}}{t(z+1)}dt=\frac{1}{z+1 }\int_{0}^{\infty} (tz+t)^{1}\cdot \sum_{n=0}^{\infty}\frac{(tz+t)^n }{n!}dt=\frac{1}{z+1 }\int_{0}^{\infty} \sum_{n=0}^{\infty}\frac{(tz+t)^{n1} }{n!}dt$. How can I continue from here ?
Last edited by skipjack; October 7th, 2019 at 04:34 AM. 
October 7th, 2019, 04:32 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 21,035 Thanks: 2271 
It seems to be $\displaystyle \ln\left(\!\frac{b+1}{a+1}\!\right)$.

October 7th, 2019, 06:11 AM  #4 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 
A fast method is : $\displaystyle \displaystyle \int_{0}^{1} \frac{x^{b}x^{a}}{\ln(x)}dx=\int_{a}^{b}[\int_{0}^{1}x^{\lambda }dx]d\lambda =\int_{a}^{b} [(1+\lambda )^{1}x^{1+\lambda }_{0}^{1} ]d\lambda =\int_{a}^{b} \frac{d\lambda }{1+\lambda }=\ln\frac{b+1}{a+1}$.


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