Integral with contour integration: $\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$

Question

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.

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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 1

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:

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

Answer 2

Let's try and hope I have not made any mistakes here. We will calculate the integral: $$\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$

Since the integrand function is even we can rewritte it as: $$\int_{-\infty}^{\infty}\frac{2x^2-1}{x^4+1}\,dx=2\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$

Considering the function $\displaystyle f(z)=\frac{2z^2-1}{z^4+1}$ we see that is has $4$ simple poles which are $\displaystyle z_1=\frac{1+i}{\sqrt{2}}, \; z_2=\frac{1-i}{\sqrt{2}}, \; z_3=\frac{-1+i}{\sqrt{2}}, \; z_4=\frac{-1-i}{\sqrt{2}}$

Only two of them lie on the upper plane (meaning they have positive imaginery part) Those are $z_1, \; z_3$.

The residues at $z_1, \; z_3$ are: $$\begin{aligned} \mathfrak{Res}\left ( f; z_1 \right ) &=\lim_{z\rightarrow z_1}(z-z_1)f(z) \\ &= \lim_{z\rightarrow \frac{1+i }{\sqrt{2}}}\left ( z-z_1 \right )\frac{2z^2-1}{(z-z_1)\left ( z-\frac{1-i}{\sqrt{2}} \right )\left ( z-\frac{-1+i}{\sqrt{2}} \right )\left ( z-\frac{-1-i}{\sqrt{2}} \right )}\\ &= \cdots\\ &= \frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\\ \end{aligned}$$

and $$\begin{aligned} \mathfrak{Res}\left ( f; z_3\right ) &=\lim_{z\rightarrow z_3}(z-z_3)f(z) \\ &= \lim_{z\rightarrow \frac{-1+i }{\sqrt{2}}}\left ( z-z_3 \right )\frac{2z^2-1}{\left ( z-\frac{1+i}{\sqrt{2}} \right )\left ( z-\frac{1-i}{\sqrt{2}} \right )\left ( z-z_3 \right )\left ( z-\frac{-1-i}{\sqrt{2}} \right )}\\ &= \cdots\\ &= -\frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\\ \end{aligned}$$

Now we are integrating on a semicircle contour with center at the origin and radius $R$. We know that $\displaystyle \oint_{\gamma}f(z)\,dz=2\pi i \left ( \frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8} -\frac{3\sqrt{2}}{8}-i\frac{\sqrt{2}}{8} \right )=\frac{\pi}{\sqrt{2}}$ .

Expanding the contour integral we have that: $$\oint_{\gamma}f(z)\,dz=\int_{-a}^{a}+\int _{{\rm arc}}$$

Letting $a \to +\infty$ we have that the arc integral goes to zero, therefore we get the value of the integral is $\dfrac{\pi}{\sqrt{2}}$ thus the original integral is $\dfrac{\pi}{2\sqrt{2}}$