How to calculate $ \int_{0}^{\infty} \frac{ x^2 \log(x) }{1 + x^4} $? I would like to calculate $$\int_{0}^{\infty} \frac{ x^2 \log(x) }{1 + x^4}$$ by means of the Residue Theorem. This is what I tried so far: We can define a path $\alpha$ that consists of half a half-circle part ($\alpha_r$) and a path connecting the first and last point of that half circle (with radius $r$) so that we have $$ \int_{-r}^{r} f(x) dx + \int_{\alpha_r} f(z) dz = \int_{\alpha} f(z) dz = 2 \pi i \sum_{v = 1}^{k} \text{Res}(f;a_v) $$ where $a_v$ are zeros of the function $\frac{x^2 \log(x) }{1+x^4}$. 
If we know $$\lim_{r \to \infty} \int_{\alpha_r} f(z) dz = 0 \tag{*} $$ then we know that $$\lim_{r \to \infty} \int_{-r}^{r} f(x) dx = \int_{-\infty}^{\infty} f(x) dx = 2 \pi i \sum_{v=1}^{k} \text{Res}(f;a_v) $$ and it becomes 'easy'. 
Q: How do we know (*) is true?

$$\underbrace{\dfrac{x^2 \log(x)}{1+x^4} dx \to \dfrac1{x^2} \dfrac{\log(1/x)}{1+1/x^4} \dfrac{-dx}{x^2}}_{x \to 1/x}$$
Hence,
\begin{align}
I = \int_0^{\infty} \dfrac{x^2 \log(x)}{1+x^4} dx & = \int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx + \int_1^{\infty} \dfrac{x^2\log(x)}{1+x^4}dx\\
& = \int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx - \int_0^1 \dfrac{\log(x)}{1+x^4}dx\\
& = \int_0^1 \dfrac{x^2-1}{1+x^4} \log(x) dx
\end{align}
\begin{align}
\int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx & = \int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{4k+2} \log(x)dx\\
& = \sum_{k=0}^{\infty}(-1)^k \int_{0}^{\infty} x^{4k+2} \log(x)dx\\
& = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(4k+3)^2}
\end{align}
\begin{align}
\int_0^1 \dfrac{\log(x)}{1+x^4} dx & = \int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{4k} \log(x)dx\\
& = \sum_{k=0}^{\infty}(-1)^k \int_{0}^{\infty} x^{4k} \log(x)dx\\
& = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(4k+1)^2}
\end{align}
$$I = \sum_{k=0}^{\infty} (-1)^k \left(\dfrac1{(4k+1)^2} - \dfrac1{(4k+3)^2}\right) = \dfrac{\pi^2}{8\sqrt2}$$
which can be obtained from the identity that the PolyLogarithmic function satisfies
$$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$
You can see the posts here and here for more details.
