The easiest way to solve the improper integral $\int_0^{+\infty} \frac{\ln(t)}{(1+t^2)^4} {d}t$?

Question

What is the easiest way to solve this improper integral?

$$I = \int_0^{+\infty} \frac{\ln(t)}{(1+t^2)^4} {d}t $$

The value of this integral according to wolfram alpha is $-\frac{23\pi}{96}$. I could find the same result but my method was too long and complicated. I'm looking for a clear and easy method.

Answer 1

Notice that the substitution $t = \tan\theta$ followed by the beta function identity gives

$$ \int_{0}^{\infty} \frac{t^{2a-1}}{(1+t^2)^b} \, dt = \int_{0}^{\infty} \sin^{2a-1} \theta \cos^{2b-2a-1}\theta \, d\theta = \frac{1}{2}\frac{\Gamma(a)\Gamma(b-a)}{\Gamma(b)} $$

which is valid for $0 < a < b$. If we plug $b = 4$, then this simplifies to

$$ \int_{0}^{\infty} \frac{t^{2a-1}}{(1+t^2)^4} \, dt = \frac{(3-a)(2-a)(1-a)\pi}{12\sin(\pi a)} $$

thanks to the Euler's reflection identity. Finally, differentiating both sides w.r.t. $a$ and plugging $a = \frac{1}{2}$ gives the desired result.

Answer 2

Using a semi-circle in the upper half plain avoiding the branch cut and using

$$f(z) = \frac{\log(z)}{(1+z^2)^4}$$

The contour

Then we have

$$\int_{-\infty}^0 \frac{\log|x|+\pi i}{(1+x^2)^4}dx +\int_{0}^\infty \frac{\log|x|}{(1+x^2)^4}dx = 2\pi i \,\mathrm{Res}(f,i)$$

By change of variable in the first integral $$\int_{0}^\infty \frac{\log|x|+\pi i}{(1+x^2)^4}dx +\int_{0}^\infty \frac{\log|x|}{(1+x^2)^4}dx = 2\pi i \,\mathrm{Res}(f,i)$$

This reduces to $$2\int_{0}^\infty \frac{\log|x|}{(1+x^2)^4}dx +\pi i \int_{0}^\infty \frac{1}{(1+x^2)^4}dx = 2\pi i \,\mathrm{Res}(f,i)$$

Note that we have a pole of order 4 at $z=i$ $$\mathrm{Res}(f,i) = \frac{23 i}{96} + \frac{5 π}{64}$$

By equating the real parts we reach the result

$$\int_{0}^\infty \frac{\log|x|}{(1+x^2)^4}dx = -\frac{23 \pi}{96}$$

Answer 3

By considering a keyhole contour, we find that

$$I_1-I_2=2\pi i(\operatorname{Res}(f,i)+\operatorname{Res}(f,-i))=\frac{23\pi^2i}{24}$$

where

\begin{align}f(z)&=\frac{\operatorname{Log}^2(z)}{(1+z^2)^4} \\I_1&=\int_0^\infty f(x)~\mathrm dx \\I_2&=\int_0^\infty f(xe^{2\pi i})~\mathrm dx \end{align}

Where we see that

$$I_1-I_2=\int_0^\infty\frac{\ln^2(z)-(\ln(z)+2\pi i)^2}{(1+z^2)^4}~\mathrm dz=\int_0^\infty\frac{4\pi^2-4\pi i\ln(z)}{(1+z^2)^4}~\mathrm dz$$

And thus,

$$I=-\frac1{4\pi}\Im(I_1-I_2)\implies I=-\frac{23\pi}{96}$$

Answer 4

$\begin{align} I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx +\int_1^{\infty}\frac{\ln x}{(1+x^2)^4}\ dx \end{align}$

Perform the change of variable $y=\frac{1}{x}$ in the second integral,

$\begin{align} I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx -\int_1^{\infty}\frac{x^6\ln x}{(1+x^2)^4}\ dx\\ &=\int_0^{1}\frac{(1-x^6)\ln x}{(1+x^2)^4}\ dx\\ &=\left[\frac{x(3x^4+5x^2+3)\ln x}{3(1+x^2)^3}\right]_0^1-\int_0^1 \frac{3x^4+5x^2+3}{3(1+x^2)^3}\ dx\\ &=-\int_0^1 \frac{3x^4+5x^2+3}{3(1+x^2)^3}\ dx\\ &=-\left[\frac{-x^3+23(1+x^2)^2\arctan x+x}{24(1+x^2)^2}\right]_0^1\\ &=-\frac{23\pi}{96}\\ \end{align}$