complex analysis - 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?

Answer

It's a bit more tricky that what you describe, but the general idea is correct. Instead of integrating from $0$ to $\infty$, one can integrate from $-\infty$ to $+\infty$ slightly above the real axis. Because of the logarithm, the integral from $-\infty$ to $0$ will give a possibly non-zero imaginary part, but the real part will be an even function of $x$. So we can write:

$$\begin{align} \int_0^{\infty}\frac{x^2\ln x}{1+x^4}dx&=\frac12\mathrm{Re}\,\int_{-\infty+i0}^{\infty+i0} \frac{x^2\ln x}{1+x^4}dx=\\&=\pi\cdot \mathrm{Re}\left[ i\left(\mathrm{res}_{x=e^{i\pi/4}}\frac{x^2\ln x}{1+x^4}+\mathrm{res}_{x=e^{3i\pi/4}}\frac{x^2\ln x}{1+x^4}\right)\right]=\\ &=\pi\cdot \mathrm{Re}\left[ i\left(\frac{\pi e^{i\pi/4}}{16}- \frac{3\pi e^{3i\pi/4}}{16}\right)\right]=\\ &=\pi\cdot\mathrm{Re}\frac{(1+2i)\pi}{8\sqrt{2}}=\frac{\pi^2}{8\sqrt{2}}. \end{align}$$

Now as far as I understand the question was about how can one justify the vanishing of the integral over the half-circle $C$ which in its turn justifies the application of residue theorem. Parameterizing that circle as $x=Re^{i\varphi}$, $\varphi\in(0,\pi)$, we see that

$$\begin{align} \int_C \frac{x^2\ln x}{1+x^4}dx=\int_0^{\pi}\frac{iR^3e^{3i\varphi}\left(i\varphi+\ln R\right)}{1+R^4e^{4i\varphi}}d\varphi=O\left(\frac{\ln R}{R}\right), \end{align}$$

which indeed vanishes as $R\rightarrow \infty$.