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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!

user1176409
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math_b
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2 Answers2

4

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. :)

Lee
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1

While Szeto answer offer a very neat solution, I was wondering why OP process lead to a wrong answer. Well turns out it's not!

We want to evaluate

$$J = \int_{0}^\infty\frac{\log^4(x)}{1+x^2}dx$$

so we introduce the function $f(z) = \frac{\log^4(z)}{1+z^2}$ for $\arg z \in \left(-\frac{\pi}{2}, \frac{3\pi}{2}\right)$ and the integration contour below

$\hskip1.5in$contour1

le $C_1$ the entire contour, using Residue Theorem we got

$$\begin{align*} \oint_{C_1}f(z)dz &= 2\pi i \text{Res}_f(i)\\ &= 2\pi i \frac{(\ln |i| + i )^4}{2i} = \frac{\pi^5}{16} \end{align*}$$

it is also true that

$$\oint_{C_1}f(z)dz = \left(\int_{\varepsilon}^{R}+\int_{-R}^{-\varepsilon}+ \int_\gamma + \int_\Gamma\right)f$$ where as OP correctly stated for $\varepsilon \to 0, R\to +\infty$, integrals along $\gamma,\Gamma$ tend to zero. We have

$$\begin{align*}\frac{\pi^5}{16} &= J + \int_{-\infty}^{0} \frac{(\ln|z|+i\pi)^4}{1+z^2}dz \quad (z=-t)\\ &= J +\int_{0}^{+\infty} \frac{(\ln|t|+i\pi)^4}{1+t^2}dt\\ &= J +\int_{0}^{+\infty} \frac{\sum_{n=0}^4\binom{4}{n}\ln^n(t)(i\pi)^{4-n}}{1+t^2}dt \end{align*}$$

taking the real part of the latter

$$2J = \frac{\pi^5}{16} + \underbrace{\int_{0}^{+\infty} \frac{6\pi^2\ln^2(t) - \pi^4}{1+t^2}dt}_{I_1}$$

we are now interested in evaluating $I_1$. It is composed by two parts

$$\begin{align*}I_1 = \int_{0}^{+\infty} \frac{6\pi^2\ln^2(t)}{1+t^2}dt- \pi^4\int_{0}^{+\infty} \frac{1}{1+t^2}dt &= 6\pi^2I_2 - \pi^4 \left [\arctan(t) \right]_0^\infty\\ &=6\pi^2I_2 -\frac{\pi^5}{2}\label{eq}\tag{$*$} \end{align*}$$

for $I_2$ we use the previously mentioned contour and we introduce $g(z) = \frac{\log^2(z)}{1+z^2}$ for $\arg z \in \left(-\frac{\pi}{2}, \frac{3\pi}{2}\right)$.

As before integrals on semicircle vanish for $\varepsilon \to 0, R\to +\infty$ and we end up with

$$\oint_{C_1}g(z)dz = -\frac{\pi^3}{4} = 2I_2 - \frac{\pi^3}{2}+2i\pi \int_{0}^{+\infty}\frac{\log t}{1+t^2}dt$$

taking real parts we can evaluate $I_2$ as

$$I_2 = -\frac{\pi^3}{8}+\frac{\pi^3}{4} = \frac{\pi^3}{8}$$

recalling $\eqref{eq}$ we conclude our calculation

$$ I_1= 6\pi^2I_2-\frac{\pi^5}{2} = \frac{\pi^5}{4}$$

And we finally came to the right answer

$$\boxed{\int_{0}^\infty\frac{\log^4(x)}{1+x^2}dx = \frac{5\pi^5}{32}}$$

Turquoise Tilt
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