Cannot solve integral

Dec 2015
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Evaluate: \(\displaystyle \int_{0}^{1} \frac{x^{b}-x^{a}}{\ln(x)}dx\;\) .
\(\displaystyle a\neq b\), \(\displaystyle a,b>0.\)
 
Dec 2015
1,085
169
Earth
Evaluate: \(\displaystyle \int_{0}^{1} \frac{x^{b}-x^{a}}{\ln(x)}dx\;\) .
\(\displaystyle a\neq b\), \(\displaystyle a,b>0.\)
The integral is similar to \(\displaystyle I(z) =\int_{0}^{1} \frac{x^z }{\ln x} dx=\frac{1}{z+1} \int_{-\infty}^{0} \frac{e^{t(z+1)}}{t(z+1)}dt=-\frac{1}{z+1 }\int_{0}^{-\infty} (tz+t)^{-1}\cdot \sum_{n=0}^{\infty}\frac{(tz+t)^n }{n!}dt=-\frac{1}{z+1 }\int_{0}^{-\infty} \sum_{n=0}^{\infty}\frac{(tz+t)^{n-1} }{n!}dt\). How can I continue from here ?
 
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skipjack

Forum Staff
Dec 2006
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It seems to be \(\displaystyle \ln\left(\!\frac{b+1}{a+1}\!\right)\).
 
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Dec 2015
1,085
169
Earth
A fast method is : \(\displaystyle \displaystyle \int_{0}^{1} \frac{x^{b}-x^{a}}{\ln(x)}dx=\int_{a}^{b}[\int_{0}^{1}x^{\lambda }dx]d\lambda =\int_{a}^{b} [(1+\lambda )^{-1}x^{1+\lambda }|_{0}^{1} ]d\lambda =\int_{a}^{b} \frac{d\lambda }{1+\lambda }=\ln|\frac{b+1}{a+1}|\).