What is wrong in my solution for $\int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(\sqrt{x^2+ \pi^2})}{\sqrt{x^2+\pi^2}} dx$

Question

Good morning! I have a doubt regarding this integral

$$ \int_{-\infty}^\infty \frac{\mathrm{arccoth}(\sqrt{x^2+ \pi^2})}{\sqrt{x^2+\pi^2}} dx$$

I have made a few attempts on solving this.

First attempt:

Let $t = \sqrt{x^2+\pi^2}$. Then the integral becomes

$$ 2 \int_\pi^\infty \frac{\mathrm{arccoth}(t)}{t}\cdot \sqrt{t^2-\pi^2} dt$$

Not quite sure what to do with this, so abandoned this attempt.

Second attempt:

Let

$$I(a) = \int_{-\infty}^\infty \frac{\mathrm{arccoth}(\sqrt{x^2+a^2})}{\sqrt{x^2+a^2}} dx$$

Then

$$I'(a) = \int_{-\infty}^\infty \frac{a}{(1-(a^2+x^2))(x^2+a^2)} dx$$

$$I'(a) = \int_{-\infty}^\infty \frac{a}{x^2+a^2}dx - \int_0^\infty \frac{a}{x^2 + a^2-1}dx$$

$$I'(a) = \pi - \frac{\pi a}{\sqrt{a^2-1}}$$

Using the fact that

$$I(\infty) = 0$$

We have

$$I(a) = \pi( a - \sqrt{a^2-1})$$

Putting $a=\pi$, I get

$$I(\pi) = \pi^2- \pi\cdot \sqrt{\pi^2-1}$$

But this is not the correct answer.

I would really appreciate it if someone can find the correct solution and point out the error in my answer. Most thankful for your efforts.

Answer 1

It seems in the second method, you have only differentiated the $\mathrm{arccoth}(\sqrt{x^2+a^2})$ part in the numerator and have missed differentiating the $\sqrt{x^2+a^2}$ down in the denominator.

There is a better way to solve this integral. Consider

$$I(a) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(a\sqrt{x^2+\pi^2})}{\sqrt{x^2+\pi^2}} dx$$

Then

$$I'(a) = \int_{-\infty}^{\infty} \frac{1}{(1-a^2\pi^2)-a^2x^2} \cdot \frac{\sqrt{x^2+\pi^2}}{\sqrt{x^2+\pi^2}} dx$$

$$I'(a) = - \frac{1}{a^2} \cdot \int_{-\infty}^{\infty} \frac{1}{x^2 + \pi^2 - \frac{1}{a^2}} dx$$

$$I'(a) = \frac{-1}{a^2} \cdot \frac{\pi}{\sqrt{\pi^2-\frac{1}{a^2}}} = \frac{-\pi}{a(\sqrt{a^2\pi^2-1})}$$

Then

$$I(a) = \int \frac{-\pi}{a(\sqrt{a^2\pi^2-1})}da = -\pi \arctan(\sqrt{a^2\pi^2-1}) + C$$

As you have observed

$$I(\infty) = 0$$

This implies that

$$I(\infty) = -\pi \cdot \frac{\pi}{2} + C = 0$$

Or

$$ C = \frac{\pi^2}{2}$$

Thus we get

$$ I(a) = \pi \cdot \left(\frac{\pi}{2} - \arctan(\sqrt{a^2\pi^2-1})\right)$$

After some rearrangement, we finally obtain a much cleaner form:

$$I(a) = \pi \cdot \arcsin \left(\frac{1}{a\pi}\right)$$

Addendum

We can easily generalize this integral for other values instead of $\pi$. We will thus have:

$$I(a,b) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(a\sqrt{x^2+b^2})}{\sqrt{x^2+b^2}} dx = \pi \cdot \arcsin \left(\frac{1}{ab}\right)$$

Which leads to some rather amusing results, such as:

$$I(1,2) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(\sqrt{x^2+4})}{\sqrt{x^2+4}}dx = \pi \cdot \arcsin(\frac{1}{2}) = \frac{\pi^2}{6} = \zeta(2)$$

And

$$I(1,1) = \int_{-\infty}^{\infty} \frac{\mathrm{arccoth}(\sqrt{x^2+1})}{\sqrt{x^2+1}}dx = \pi \cdot \arcsin(1) = \frac{\pi^2}{2}$$