Evaluate $\displaystyle \int_{0}^{\infty} \frac{\ln x}{(x+a)(x+b)} \ dx \ \ a,b>0 \ (a \ne b) $

Unlike the last integral, try not to use contour integration.

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- Jun 13th 2011, 08:57 PMRandom Variableimproper integral #4
Evaluate $\displaystyle \int_{0}^{\infty} \frac{\ln x}{(x+a)(x+b)} \ dx \ \ a,b>0 \ (a \ne b) $

Unlike the last integral, try not to use contour integration. - Jun 14th 2011, 02:38 PMDrexel28Re: improper integral #4
- Jun 14th 2011, 04:30 PMRandom VariableRe: improper integral #4
There is an easier approach.

Make the substitution $\displaystyle u = \frac{ab}{x} $ and see what happens. - Jun 14th 2011, 04:58 PMDrexel28Re: improper integral #4
Yes, that may work. But it wasn't what was most natural to me. I evaluated the integral on Wolfram Alpha for a couple of specific values and noticed that it was always of the form $\displaystyle F(b)-F(a)$ where $\displaystyle b>a$ that just screamed use a double integral to me.

- Jun 14th 2011, 05:48 PMRandom VariableRe: improper integral #4
The substitution is not something I came up with. My automatic approach would be to find the Cauchy principal value using contour integration and then drop the P.V. label because the integral is convergent.

But if you make that substitution you get

$\displaystyle \int^{\infty}_{0} \frac{ \ln x}{(x+a)(x+b)} \ dx = \ln (ab) \int^{\infty}_{0} \frac{du}{(u+a)(u+b)} - \int^{\infty}_{0} \frac{ \ln u}{(u+a)(u+b)} \ du$

so $\displaystyle \int^{\infty}_{0} \frac{\ln x}{(x+a)(x+b)} \ dx = \frac{\ln (ab)}{2} \int^{\infty}_{0} \frac{du}{(x+a)(x+b)} \ dx $

EDIT: $\displaystyle = \frac{\ln a + \ln b}{2(a-b)} \int^{\infty}_{0} \Big(\frac{1}{x+b} - \frac{1}{x+a} \Big) \ dx = \frac{\ln a + \ln b}{2(a-b)} \ (\ln a - \ln b) $

$\displaystyle = \frac{\ln^{2} a - \ln^{2} b}{2(a-b)} $ - Jun 14th 2011, 07:41 PMRandom VariableRe: improper integral #4
The same substitution can be used to evaluate $\displaystyle \int^{b}_{a} \frac{\ln x}{(x+a)(x+b)} \ dx $.