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 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
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 Originally Posted by idontknow 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 ?

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 $\displaystyle \ln\left(\!\frac{b+1}{a+1}\!\right)$. Thanks from idontknow October 7th, 2019, 06:11 AM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 A fast method is : $\displaystyle \displaystyle \int_{0}^{1} \frac{x^{b}-x^{a}}{\ln(x)}dx=\int_{a}^{b}[\int_{0}^{1}x^{\lambda }dx]d\lambda =\int_{a}^{b} [(1+\lambda )^{-1}x^{1+\lambda }|_{0}^{1} ]d\lambda =\int_{a}^{b} \frac{d\lambda }{1+\lambda }=\ln|\frac{b+1}{a+1}|$. Tags integral, solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post amirstark Complex Analysis 1 November 23rd, 2015 04:27 PM Rosa Calculus 4 September 27th, 2015 05:31 PM ghafarimahsa Calculus 1 September 11th, 2013 12:07 PM Ad van der ven Calculus 3 December 17th, 2011 09:53 AM Niko Bellic Calculus 3 December 1st, 2008 07:47 AM

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