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