MHB Sum involving pi, ln(2) and Catalan's constant

[Sherlock1]

Inspired by http://www.mathhelpboards.com/showthread.php?560-integrals thread: By considering the integral $\displaystyle \int_{0}^{\infty} \frac{\ln(1+x^2)}{1+x^2}\;{dx}$ or otherwise,
show that $\displaystyle \sum_{k \ge 0}~\sum_{0 \le j \le k}\frac{(-1)^k}{(2k+3)(j+1)} = \frac{1}{2}\pi\ln(2)-G$ where $G$ is Catalan's constant.

 

[chisigma]

In another post it has been demonstrated that...

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

Now with simple steps You can find that...

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

... so that is...

$\displaystyle \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x^{2}}\ dx = \frac{\pi}{2}\ \ln 2 + \int_{0}^{1} \frac{\ln x}{1+x^{2}}\ dx $ (3)

Some years ago I 'discovered' that is...

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

... so that is...

$\displaystyle \int_{0}^{1} \frac{\ln x}{1+x^{2}}\ dx = - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = - G$ (5)

... and finally...

$\displaystyle \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x^{2}}\ dx = \frac{\pi}{2}\ \ln 2 -G$ (6)

Now find the series expansion of the definite integral in (6) is perfectly possible... but in my opinion is more elegant the (6) without any more 'processing'...

Kind regards

$\chi$ $\sigma$

 

[Sherlock1]

Good job. Here's the series expansion done just for completion.

$\begin{aligned} \int_{0}^{1}\frac{\ln(1+x^2)}{1+x^2}\;{dx} & = \int_{0}^{1}\bigg(\sum_{k \ge 0}(-1)^kx^{2k}\bigg)\bigg(\sum_{k\ge 0}\frac{(-1)^kx^{2k+2}}{k+1}\bigg)\;{dx} \\& = \int_{0}^{1}\sum_{k \ge 0}~\sum_{0 \le j \le k} (-1)^{k-j}x^{2k-2j}\cdot\frac{(-1)^jx^{2j+2}}{j+1}\;{dx} \\& = \int_{0}^{1}\sum_{k \ge 0}~\sum_{0 \le j \le k} \frac{(-1)^k x^{2k+2}}{j+1}\;{dx} \\& = \sum_{k \ge 0}~\sum_{0 \le j \le k} ~\int_{0}^{1}\frac{(-1)^k x^{2k+2}}{j+1}\;{dx} \\& = \sum_{k \ge 0}~\sum_{0 \le j \le k} \frac{(-1)^k}{(j+1)(2k+3)}.\end{aligned}$