# Integral of arctan and log using contour integration

$$\int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx = \frac{\pi^3}{12}+\pi \log^2(2)$$

Lemma

$$\int^\infty_0 \frac{\log^3(1 + x^2)}{x^2}\,dx = \pi^3+ 3 \pi \log^2(4)$$

Start by the following

$$\int^{\infty}_0 x^{-p}(1+x)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)}$$

Let $x\to x^2$

$$\int^{\infty}_0 x^{-2p+1}(1+x^2)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{2\Gamma(1-s)}$$

Let $p = 3/2$
$$\int^{\infty}_0 \frac{1}{x^2(1+x^2)^{1-s}} dx= \frac{\Gamma(-1/2)\Gamma(3/2-s)}{2\Gamma(1-s)}$$

By differentiation

$$\int^\infty_0 \frac{\log^3(1+x^2)}{x^2}\,dx = \frac{\Gamma(-1/2)}{2} \left [ \frac{d^3}{ds^3}\frac{\Gamma(3/2-s)}{2\Gamma(1-s)} \right ]_{s = 1} = \pi^3+ 3 \pi \log^2(4)$$

Now, Consider the function

$$f(z) = \frac{\log^3(1-iz)}{z^2}$$

Define the principle logarithm as follows

$$\log z = \log|z| + i\mathrm{Arg}(z) \,\,\,$$

Note that for $x > 0$ the argument can be evaluated as

$$\log z = \log\sqrt{x^2+y^2} + i\arctan(y/x)$$
For the principle logarithm the branch cut is defined as $\Im(1-iz) =0 ,\Re(1-iz) \leq 0$ which then reduces for $z = x+iy$ to $x=0 , 1+ y \leq 0$ . Hence the branch cut on the imaginary axis where $x=0 , y < -1$ . Also note that $f(z)$  is analytic on the punctured plane since around $z =0$

$$\frac{\log^3(1-iz)}{z^2} = -\frac{((iz)+(iz)^2/2+(iz)^3/3 +\mathcal{O}(z^4))^3}{z^2} =\mathcal{O}(z^2)$$

Hence at $z = 0$ we have a removable singularity. Define the following contour were we avoid the branch cut

$$\int_{C_R}f(z)\,dz+\int_{-R}^{0}\frac{\left(\log(\sqrt{1+x^2})+i\arctan(x)\right)^3}{x^2}dx + \int_{0}^R\frac{(\log(\sqrt{1+x^2})+i\arctan(x))^3}{x^2}dx= 0$$
$$\int_{C_R}f(z)\,dz+\int_{0}^{R}\frac{\left(\log(\sqrt{1+x^2})-i\arctan(x)\right)^3}{x^2}dx + \int_{0}^R\frac{(\log(\sqrt{1+x^2})+i\arctan(x))^3}{x^2}dx= 0$$

Note that

$$(x+y)^3 = x^3+y^3+3x^2y+3yx^2$$

This simplifies to

$$\int_{C_R}f(z)\,dz+\int_{0}^{R}\frac{-3 \arctan^2(x) \log(1 + x^2) + 1/4 \log^3(1 + x^2)}{x^2}\,dx= 0$$

Taking the limit

$$\left|\int_{C_R}f(z)\,dz \right| = \left|\int_{C_R}\frac{\log^3(1-iz)}{z^2}\,dz \right| \leq \pi R \frac{(\log |1+R| + 2\pi )^3}{R^2} \sim_{\infty} 0$$

So this simplifies to

$$\int_{0}^{\infty}\frac{ \arctan^2(x) \log(1 + x^2)}{x^2} \,dx = \frac{1}{12} \int^\infty_0 \frac{\log^3(1 + x^2)}{x^2}\,dx$$

Using the Lemma we reach our result.

This entry was posted in Beta function, Contour Integration and tagged , , , , , , , , . Bookmark the permalink.

### 4 Responses to Integral of arctan and log using contour integration

1. tired says:

i see my comments are fruitful.. 🙂

• Zaidalyafeai says:

Yeah, your comments are better than many answers. What brought you here ?

2. tired says:

your pdf-book was recommended somewhere on mse which somehow hinted me in the direction of your nice blog

• Zaidalyafeai says:

Glad you found it nice.