# Catalan's Constant

• Aug 11th 2010, 05:56 PM
chiph588@
Catalan's Constant
Show $\displaystyle \displaystyle \int_1^\infty \frac{\log x}{x^2+1}dx = G,$ where $\displaystyle \displaystyle G$ is Catalan's constant.
• Aug 11th 2010, 08:16 PM
chisigma
Setting $\displaystyle u=\frac{1}{x}$ the integral becomes...

$\displaystyle \displaystyle \int_{1}^{\infty} \frac{\ln x}{1+ x^{2}}\ dx = - \int_{0}^{1} \frac{\ln u}{1 + u^{2}}\ du$ (1)

Now with a little of patience You can demonstrated that in general is...

$\displaystyle \displaystyle \int_{0}^{1} u^{n}\ \ln u\ du = -\frac{1}{(n+1)^{2}}$ (2)

... so that, tacking into account that for $\displaystyle |u|<1$ is...

$\displaystyle \displaystyle \frac{1}{1+ u^{2}} = \sum_{n=0}^{\infty} (-1)^{n}\ u^{2n}$ (3)

is...

$\displaystyle \displaystyle \int_{1}^{\infty} \frac{\ln x}{1+ x^{2}}\ dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}}$ (4)

This integral is very similar to the integral resolved here...

http://www.mathhelpforum.com/math-he...ls-151443.html

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$