# Integrate $\int_a^\infty\frac{\sqrt{x^2-a^2}}{\sinh x}\,dx$

I wonder whether the following integral ($$a\in\mathbb{R}$$, $$a>0$$) $$S(a)=\int_a^\infty\frac{\sqrt{x^2-a^2}}{\sinh x}\,dx$$ admits a closed form (perhaps using some known special functions).

The integral representation of $$K_1(z)$$ gives just $$S(a)=2a\sum\limits_{n=0}^{\infty}\frac{K_1\big((2n+1)a\big)}{2n+1}$$ which doesn't lead me to anything meaningful.

An alternative form comes from contour integration: $$\frac{S(a\pi)}{\pi^2}=\frac{1-2a}{4}+\sum_{n=1}^{\infty}(-1)^{n-1}(\sqrt{n^2+a^2}-n).$$

For a context, this is what I arrive at in this answer.

• Mathematica doesn't provide an answer; I imagine this one would be quite difficult. – Adrian Keister Jul 8 '19 at 19:06

Let's denote:

$$F(a)=\frac{1}{2}+\frac{d}{da} \frac{S(a\pi)}{\pi^2} \tag{1}$$

Remember for later that $$S(0)= \frac{\pi^2}{4}$$.

From the second expression in the OP we obtain:

$$F(a)=a \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n^2+a^2}} \tag{2}$$

In this answer I found that:

$$\frac{F(a)}{a}=\log 2 +\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2} \right)^{2k} \tag{3}$$

and

$$\frac{F(a)}{a}=\log 2 +\int_0^\infty \frac{J_0 (a x)-J_0 (a x/2)}{e^x-1}dx \tag{4}$$

Integrating (1) from $$0$$ to $$a$$, we have:

$$\int_0^a F(a') da'=\frac{a}{2}+\frac{S(a\pi)}{\pi^2}-\frac{1}{4}$$

On the other hand (3) gives us:

$$\int_0^a F(a') da'=\frac{\log 2}{2} a^2+2\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2 (k+1)}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2} \right)^{2k+2}$$

This gives us explicit Taylor series for $$S(a)$$:

$$S(a)= \frac{\pi^2}{4}-\frac{\pi a}{2}+\frac{\log 2}{2} a^2+ \frac{a^2}{2} \sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2 (k+1)}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2 \pi} \right)^{2k} \tag{5}$$

While this is not a closed form, it still may prove useful.

From (4) we obtain:

$$\int_0^a F(a') da'=\frac{\log 2}{2} a^2+a \int_0^\infty \frac{J_1 (a x)-2 J_1 (a x/2)}{x(e^x-1)}dx$$

Which gives us an additional integral form for the function:

$$S(a)= \frac{\pi^2}{4}-\frac{\pi a}{2}+\frac{\log 2}{2} a^2+ \pi a \int_0^\infty \frac{J_1 (a x)-2 J_1 (a x/2)}{x(e^{\pi x}-1)}dx \tag{6}$$

• Thanks! BTW, $(6)$ simplifies to $S(a)=\displaystyle\frac{\pi^2}{4}-\frac{\pi a}{2}+\pi a\int_0^\infty\frac{J_1(ax)\,dx}{x(e^{\pi x}+1)}$. – metamorphy Jul 17 '19 at 12:54
• @metamorphy, thank you, I see that now. That makes the integral much nicer to numerically compute – Yuriy S Jul 17 '19 at 13:00