complex analysis - Integral with contour integration

I want to evaluate the integral:
$$\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$

using contour integration.

I re-wrote it as: $\displaystyle \int_{0}^{\infty}\frac{2x^2-1}{x^4+1}\,dx$. I am considering of integrating on a semicircle contour with center at the origin. I considered the function $\displaystyle f(z)=\frac{2z^2-1}{z^4+1}$ which has $4$ simple poles but only two of them lie on the upper half plane and included in the contour which are: $\displaystyle z_1=\frac{1+i}{\sqrt{2}}, \;\; z_2=\frac{-1+i}{\sqrt{2}}$.

The residue at $\displaystyle z_1$ equals $\displaystyle \mathfrak{Res}\left ( f; z_1 \right )=-\frac{2i-1}{2\sqrt{2}}$ while the residue at $z_2$ equals $\displaystyle -2\sqrt{2}i-2\sqrt{2}$. (if I have done the calculations right)

Now, I don't know how to continue. Should I find the residues at the other poles as well and the say $\displaystyle \oint_{C}f(z)=2\pi i \sum res$ where $C$ is the semicircle contour and then expand it? That is:

$$\oint_{C}f(z)\,dz=\int_{0}^{a} + \int_{{\rm arc}}$$

Then let $a \to +\infty$ then than arc integral would go to zero.
But I don't know how to proceed.

---

I had dealt with this integral with residues converting it into a minus infinity to infinity integral but with contours I am having a bit of problem.

Therefore I'd like some help.

Answer

You could also use the following contour.

$$\int_{-\infty}^0 \frac{2x^2-1}{x^4+1}\operatorname dx =\int_{0}^{+\infty} \frac{2x^2-1}{x^4+1}\operatorname dx $$

Has an analytic continuation as $$\int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz $$ with 4 poles, but just on pole inside of the contour.

$$\operatorname*{res}_{z=e^{i\frac{\pi}{4}}} f(z) = \frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8} $$

I think you made a calculation error here (?)

Using $$\int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz = \color{blue}{\int_{\Gamma_1}\frac{2z^2-1}{z^4+1}\operatorname dz} + \int_{\Gamma_2}\frac{2z^2-1}{z^4+1}\operatorname dz + {\color{red}{\int_{\Gamma_3}\frac{2z^2-1}{z^4+1}\operatorname dz}}$$

• Now $\color{blue}{\int_{\Gamma_1} \to 0}$ as $R\to +\infty$ which can be proven using the triangle inequality.


• Use $\Gamma_2 \leftrightarrow z(x) = x$ and $x:0\to R$


• Use $\color{red}{\Gamma_3 \leftrightarrow z(y) = iy}$ and $y: R \to 0$

Which finally results in (as $R \to +\infty$)

$$2\pi i \left(\frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\right) = \color{blue}{0}+\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx + \color{red}{ i \int_{+\infty}^0\frac{-2y^2-1}{y^4+1}\operatorname dy}$$

Here you can read the real parts which results in:

$$\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx = \frac{\pi\sqrt{2}}{4}$$