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logarithmic Integral

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Post Fri Jul 01, 2016 10:01 am

Posts: 82
\(\displaystyle \displaystyle \int_{0}^{1}\frac{\ln(1+x^2)}{1+x}dx\)

Post Sun Jul 03, 2016 2:33 pm

Posts: 27
Define $f(a) = \int_0^1 \frac{\log{(1+ax^2)}}{1+x}\text{d}x$

We want to find the value of $f(1)$.

Take note that $f(0) = 0$. This will come in handy later.

Differentiate under the integral sign with respect to $a$.

$f'(a) = \int_0^1 \frac{x^2}{(1+ax^2)(1+x)} \text{d}x$

Evaluate the integral.

$f'(a) = -\frac{\tan^{-1}{\sqrt{a}}}{\sqrt{a}(1+a)} + \frac{\log{2}}{1+a} + \frac{\log{(1+a)}}{2a} - \frac{\log{(1+a)}}{2(1+a)}$

Now, since $f(0) = 0$, by the Fundamental Theorem of Calculus, we can find the value of $f(1)$ without finding $f(a)$ in general, allowing us to avoid the Dilogarithm, which is part of the general expression for $f(a)$.

The value of $f(1)$ is therefore $\int_0^1 f'(a) \text{d}a = \int_0^1 -\frac{\tan^{-1}{\sqrt{a}}}{\sqrt{a}(1+a)} + \frac{\log{2}}{1+a} + \frac{\log{(1+a)}}{2a} - \frac{\log{(1+a)}}{2(1+a)} \text{d}a$

The third term is one quarter of $\zeta(2)$ and the other terms are subject to elementary integration.

So $f(1) = -\frac{\pi^2}{16} + \log^2{2}+ \frac{\pi^2}{24} - \frac{\log^2{2}}{4} = \frac{3\log^2{2}}{4}-\frac{\pi^2}{48}$

Post Mon Sep 12, 2016 7:04 pm

Posts: 82
Thanks Paradoxica.

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