Given the following integral, how would you solve it?
$$
\int_0^\infty{\frac{dx}{x^{\frac{2}{5}} (x^2 + 1)^2 }}
$$
because $f(x)$ is an even function:
$$
\int_{-\infty}^\infty{\frac{dx}{2x^{\frac{2}{5}} (x^2 + 1)^2 }}
$$
By Residue integration, we have:
$$
\int_\Gamma f(z)dz = 2 \pi i \sum \text{Res} \{f, z_i \}
$$
The countour $\Gamma$ is:
$$
\Gamma = [\varepsilon, R] \cup\gamma_{R}(0, \pi) \cup [-R, -\varepsilon] \cup \gamma_\varepsilon (-\pi, 0)
$$
where:
$$
\gamma_r(a, b): \; z = re^{it} \; \; \; \; t\in (a, b)
$$
hence:
$$
\text{Res} \{f, z = 0 \} = \lim_{z \to 0} \; \; (z – 0) \frac{1}{2z^{\frac{2}{5}} (z^2 + 1)^2 } = 0.
$$
$$
\text{Res} \{f, z = i \} = \lim_{z \to i} \; \; \frac{d}{dz} \left[(z – i)^2\frac{1}{2z^{\frac{2}{5}} (z^2 + 1)^2 } \right] = -\frac{7}{40}e^{3i \pi / 10}
$$
$$
\int_\Gamma f(z)dz = -\frac{7 i \pi}{20}e^{3i \pi / 10}
$$
Now, splitting the integrals:
$$
\int_\Gamma f(z)dz = \int_{\gamma_R} f(z)dz + \int_{\gamma_\varepsilon} f(z)dz + \int_{-R}^{-\varepsilon} f(z) dz + \int^{R}_{\varepsilon} f(z) dz
$$
$$
\int_{\gamma_\varepsilon} f(z)dz = \pi i \text{Res} \{f, z = 0 \} = 0
$$
performing the limit where $R \to \infty$ and $\varepsilon \to 0$:
$$
\int_\Gamma f(z)dz = \int_{-\infty}^{\infty} f(z)dz = -\frac{7 i \pi}{20}e^{3i \pi / 10}
$$
But this is clearly wrong. I'm doing it right? How should I solve the integral with residues?
Best Answer
The problems started at the very beginning, when you claimed the function under the integral is even. Not only it's not even, but note that as a real function it is not even well defined on the negative real axis. You can only define it as a complex function, and that requires choosing a branch.
Your choice of contour is good, and the complex function is $f(z)=\frac{1}{z^{\frac{2}{5}}(z^2+1)^2}$, where the chosen branch uses the argument between $-\frac{\pi}{2}$ and $\frac{3\pi}{2}$. The only difficulty is to understand the integral on the segment $[-R,-\epsilon]$. For each $t$ in this interval we have, by definition:
$t^{\frac{2}{5}}=e^{\frac{2}{5}\log(t)}=e^{\frac{2}{5}(ln|t|+i\pi)}=(-t)^{\frac{2}{5}}e^{i\pi\frac{2}{5}}.$
That is because $|t|=-t$, and the argument of $t$ on the chosen branch is $\pi$. So after a change of variable, the integral becomes:
$\int_{-R}^{-\epsilon}\frac{e^{-i\pi\frac{2}{5}}dt}{(-t)^{\frac{2}{5}}(t^2+1)^2}=e^{-i\pi\frac{2}{5}}\int_{\epsilon}^R\frac{dt}{t^{\frac{2}{5}}(t^2+1)^2}\to e^{-i\pi\frac{2}{5}}\int_0^{\infty}\frac{dt}{t^{\frac{2}{5}}(t^2+1)^2}.$
So all together, the integral over $\Gamma$ tends to $(1+e^{-i\pi\frac{2}{5}})\int_0^{\infty}f(t)dt$. Now compare with what the residue theorem gives you.