I am trying to evaluate $$I=\int_0^\infty\frac{1}{(\sqrt{x})^3+1}dx$$ I know that I can probably do a change of variable and then split it to partial fraction, which should be integrable without any complex analysis. However, I am struggling with solving this integral with a contour. For example, I tried using the semi-circle in the upper half-plane, but I cannot figure out $\int_{-\infty}^0\frac{1}{(\sqrt{x})^3+1}dx$ as a multiple of $I$. I changed it to $\int_\infty^0\frac{1}{(\sqrt{e^{i\pi}x})^3+1}dx$, and do not know how to proceed. Is partial fraction here inevitable?
Integration using a contour
complex-analysiscontour-integration
Related Solutions
Note that $$ f(z) = \frac{e^{iz}}{z-i} = \frac{(z+i)e^{iz}}{(z+i)(z-i)} = \frac{(z+i)e^{iz}}{z^2+1}. $$
So, if $z=x$ is real, then $$ f(x) = \frac{(x+i)(\cos x + i \sin x)}{x^2+1} = \frac{x\cos x - \sin x + i(\cos x + x\sin x)}{x^2+1}. $$
Let $\Gamma$ be the closed curve consisting of the interval $[-R,R]$ together with the semi-circle $C : z=Re^{it}$, $0 < t < \pi$ (for $R > 1$). By the residue theorem,
$$ \int_{\Gamma} f(z)\,dz = \int_{[-R,R]} f(z)\,dz + \int_C f(z)\,dz = 2\pi i \operatorname{Res}(f;z=i) $$
(since $z=i$ is the only singularity of $f$ inside $\Gamma$). Finally, let $R\to\infty$.
The integral over $C$ will tend to $0$ by Jordan's lemma, and you will end up with
$$ \int_{-\infty}^\infty \frac{x\cos x - \sin x}{x^2+1}\,dx + i \int_{-\infty}^\infty \frac{x\sin x + \cos x}{x^2+1}\,dx = 2\pi i \operatorname{Res}(f;z=i). $$
To finish off, compute the residue and look at the imaginary part of the equality above.
Feynman isn't necessarily antangonistic to contour integration. They sometimes work together well.
$$J=\int^\infty_0\frac{\log^4(x)}{1+x^2}dx=I^{(4)}(0)$$
where $$I(a)=\int^\infty_{0}\frac{x^{a}}{1+x^2}dx$$ with $-1<a<1$.
Take $C$ as a keyhole contour, centered at the origin, avoiding the positive real axis.
Let $f(z)=z^a(1+z^2)^{-1}$ with branch cut on the positive real axis, implying $z^a\equiv \exp(a(\ln|z|+i\arg z))$ where $\arg z\in[0,2\pi)$.
Firstly, by residue theorem, $$\oint_C f(z)dz=2\pi i\bigg(\operatorname*{Res}_{z=i}f(z)+\operatorname*{Res}_{z=-i}f(z)\bigg)$$
We have $$\operatorname*{Res}_{z=i}f(z)=\frac{\exp(a(\ln|i|+i\arg i))}{i+i}=\frac{e^{\pi ia/2}}{2i}$$ $$\operatorname*{Res}_{z=-i}f(z)=\frac{\exp(a(\ln|-i|+i\arg -i))}{-i-i}=-\frac{e^{3\pi ia/2}}{2i}$$
Thus, $$\oint_C f(z)dz=\pi(e^{\pi ia/2}-e^{3\pi ia/2})$$
On the other hand, $$\oint f(z)dz=K_1+K_2+K_3+K_4$$
where
$$ K_1=\lim_{R\to\infty}\int^{2\pi}_0 f(Re^{it})iRe^{it}dt =\lim_{R\to\infty}2\pi f(Re^{ic})iRe^{ic}=0 \qquad{c\in[0,2\pi]}$$
$$K_2=\lim_{r\to0^+}\int_{2\pi}^0 f(re^{it})ire^{it}dt =\lim_{r\to0^+}2\pi f(re^{ic})ire^{ic}=0 \qquad{c\in[0,2\pi]}$$
$$K_3=\int^\infty_0 f(te^{i0})dt=\int^\infty_0\frac{e^{i0}t^a}{t^2+1}dt=I$$
$$K_4=\int_\infty^0 f(te^{i2\pi})dt=-\int^\infty_0\frac{e^{2\pi ia}t^a}{t^2+1}dt=-e^{2\pi ia}I$$
For $K_1,K_2$, please respectively note the asymptotics $f(z)\sim z^{a}$ for small $|z|$ and $f(z)=O(z^{a-2})$ for large $|z|$.
Therefore, $$I-e^{2\pi ia}I=\pi(e^{\pi ia/2}-e^{3\pi ia/2})$$ $$\implies I=\pi\frac{e^{\pi ia/2}-e^{3\pi ia/2}}{1-e^{2\pi ia}} =\pi\frac{e^{-\pi ia/2}-e^{\pi ia/2}}{e^{-\pi i a}-e^{\pi ia}} =\pi\frac{\sin(\pi a/2)}{\sin(\pi a)} =\frac{\pi}2\sec\left(\frac{\pi a}2\right) $$
Let $T=\tan(\pi a/2)$, $S=\sec(\pi a/2)$. $$I^{(4)}(a)=\frac{\pi}2\frac{\pi^4(T^4+18S^2T^2+5S^4)}{16}$$
Hence, $$J=I^{(4)}(0)=\frac{\pi}2\frac{\pi^4(0+0+5\cdot 1)}{16}=\color{red}{\frac{5\pi^5}{32}}$$ The tedious differentiation is done by calculator. :)
Best Answer
Hint
Replace $x=u^2$ to obtain $$ I=\int_0^\infty \frac{2udu}{u^3+1} $$ and perform contour integration on the above expression.