I was pondering this integral a little today. I see from the MathWorld site that the solution is
\frac{\text{Catalan}}{2}+\frac{\pi}{8}ln(2)
I think if we are clever enough, we may be able to transform the integral into some other integrals that can be done. Though not by elementary means.
It can be shown that \int_{0}^{1}\frac{tan^{-1}(x)}{x}dx=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}=\text{Catalan}
Also, it can be shown that \int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{\pi}{8}ln(2)
Can we do some manipulations and transform our integral into this:
\frac{1}{2}\int_{0}^{1}\frac{tan^{-1}(x)}{x}dx+\int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{K}{2}+\frac{\pi}{8}ln(2)
I will have to look at it some more tomorrow. Fun and challenging problem.
I am certain a lot of you out there are certainly better than me at this, so what do you think?.
It is there. I can smell it. I see many different approaches and identities, but am missing something.
I also tried the series for arctan: \sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}
If we use that with the rest of our integral, we get:
\int_{0}^{1}\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(x^{2}+1)(2k+1)}
Now, some manipulations here and there and it looks like it should fall into place.
There is uniform convergence from 0 to 1, so we can switch our signs around.
\int_{0}^{1}\frac{x^{2k}}{x^{2}+1}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}
If we integrate \int x^{2k}dx, we get \frac{x^{2k+1}}{2k+1}.
Which when multiplied with the other 2k+1, we get the catalan constant.
Like I said, it is right there, but.....
I am just throwing some things out there. Having fun with the problem.
