My Math Forum Cannot solve integral

 Calculus Calculus Math Forum

 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

Quote:
 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top