I am trying to solve the following integral using a contour (large semi-circle connected to smaller semi-circle in the upper-half plane):
$$\int_0^{\infty} \frac{\log^4(x)}{1+x^2} dx.$$
I have split the contour into 4 parts – the large semi-circle, the small semi-circle, the part on the negative real axis and the part on the positive real axis.
The integral of the function over the contour is $2\pi i \sum Res(f)$, which is $\pi^5$. The function has poles: $i$ and $-i$, each of order $1$, but $i$ is the only pole contained in the contour.
The integral over the large semi-circle is $0$ as the large radius approaches infinity and the integral over the small semi-circle is $0$ as the small radius approaches $0$.
I take the real part of both sides and the following is left:
$$
\pi^5 = 2\int_0^{\infty} \frac{\log^4(x)}{1+x^2} dx
+ \int_{-\infty}^0 \frac{-6\pi^2\log^2(x) + \pi^4}{1+x^2} dx
$$
My final answer is $5\pi^5/8$, but the correct answer is $5\pi^5/32$.
Any suggestions? Thank you!
Best Answer
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. :)