Is there a rapider or more elegant way to evaluate $\int_0^{+\infty} \frac{\cos(\pi x)\ \text{d}x}{e^{2\pi \sqrt{x}}-1}$?

Question

$$\int_0^{+\infty} \frac{\cos(\pi x)\ \text{d}x}{e^{2\pi \sqrt{x}}-1}$$

First attempt

• $x\to t^2$

• Geometric series by writing the denominator as $e^{2\pi t}(1 - e^{-2\pi t})$

• $\cos(\pi t^2) = \Re e^{i\pi t^2}$

This leads me to

$$2\sum_{k = 0}^{+\infty} \int_0^{+\infty} t e^{i\pi t^2}e^{-\alpha t}\ \text{d}t$$

Where $\alpha = 2\pi (k+1)$.

Now I thought about writing it again as

$$-2\sum_{k = 0}^{+\infty}\frac{d}{d\alpha} \int_0^{+\infty} e^{i\pi t^2}e^{-\alpha t}\ \text{d}t$$

The last integral can be evaluated with the use of the Imaginary Error Function, hence a Special Function method.

Yet it doesn't seem me the best way.

Second Attempt

Basically like the previous one with the difference that

• $\cos( \cdot )$ stays as it;

• $\pi t^2 \to z$;

And this brings

$$-\frac{1}{\sqrt{\pi}}\frac{d}{d\alpha} \sum_{k = 0}^{+\infty}\int_0^{+\infty} \frac{\cos(z)}{z} e^{-\alpha \sqrt{\frac{z}{\pi}}}\ \text{d}z$$

But in both cases what I am thinking are just numerical methods. Or at least I could give a try with the stationary phase but... meh.

I don't know if I can use residues for this, actually. Even if taking a look at the initial integral, there is this additional way:

$$\frac{1}{e^{2\pi t} -1} = \frac{1}{(e^{\pi t}+1)(e^{\pi t}-1)}$$

Which for example has a pole at $t = +i$...

But using residues I would obtain

$$\pi \cos(\pi)$$

Where as the correct numerical result (which I checked with Mathematica) is

$$\color{blue}{0.0732233(...)}$$

And it seems there is not a closed form for this.

Any hint/help?

Answer 1

$$\color{blue}{\int_0^\infty {\frac{{\cos (\pi x)}}{{{e^{2\pi \sqrt x }} - 1}}dx} = \frac{{2 - \sqrt 2 }}{8}}$$

Simple manipulation gives $$\int_0^\infty {\frac{{\cos (\pi x)}}{{{e^{2\pi \sqrt x }} - 1}}dx} = 2\int_0^\infty {\frac{{x\cos (\pi {x^2})}}{{{e^{2\pi x}} - 1}}dx} = \frac{1}{2} \int_0^\infty {\frac{{\sin (\pi {x^2})}}{{{{\sinh }^2}\pi x}}dx} $$ we evaluate the last integral.

---

Consider the function $$f(z) = \frac{{{e^{i\pi {z^2}}}{e^{2\pi z}}}}{{{{\sinh }^2}\pi z\sinh 2\pi z}}$$ note that $$f(z) - f(z + 2i) = 2\frac{{{e^{i\pi {z^2}}}}}{{{{\sinh }^2}\pi z}}$$

Integrate $f(z)$ around the rectangle with vertices $-R, R, R+2i, -R+2i$, with semicircle indentations at $0$ and $2i$. The indentation at $0$ is above the $x$-axis, while indentation at $2i$ is also above the line $\Im(z) = 2$, denote these two circles as $C_1$ and $C_2$ respectively, both with radius $r$.

Then $$\tag{1} 2\pi i \sum_{k=1}^4 \text{Res}[f(z),\frac{k}{2}i] = 2 \mathcal{P}\mathcal{V} \int_{ - \infty }^\infty {\frac{{{e^{i\pi {z^2}}}}}{{{{\sinh }^2}\pi z}}dz} + \int_{C_1} f(z) dz + \int_{C_2} f(z) dz $$

As $r\to 0$, $$\begin{aligned} \int_{C_1} f(z) dz + \int_{C_2} f(z) dz &= - \int_0^\pi {ri{e^{ix}}\left[ {f(r{e^{ix}}) - f(2i + r{e^{ix}})} \right]dx} \\ &= - \int_0^\pi {\frac{2}{{{\pi ^2}{r^2}{e^{2ix}}}}ri{e^{ix}}dx} + o(1) \\ &= - \frac{4}{{{\pi ^2}r}} + o(1) \end{aligned}$$ where we used the expansion of $f(z)-f(z+2i)$ around $0$. Note the dominant term is real.

Thus $(1)$ becomes $$2\pi i\left( {\frac{1}{\pi } - \frac{1}{{\sqrt 2 \pi }} + \frac{i}{{\sqrt 2 \pi }}} \right) = 2\mathcal{P}\mathcal{V}\int_{ - \infty }^\infty {\frac{{{e^{i\pi {z^2}}}}}{{{{\sinh }^2}\pi z}}dz} - \frac{4}{{{\pi ^2}r}} + o(1)$$

Comparing imaginary part gives $$\int_0^\infty {\frac{{\sin (\pi {x^2})}}{{{{\sinh }^2}\pi x}}dx} = \frac{{2 - \sqrt 2 }}{4} $$

---

Denote $$I(a) = \int_0^\infty {\frac{{\cos (a\pi x)}}{{{e^{2\pi \sqrt x }} - 1}}dx} $$ then additional values include

$$\begin{aligned}I(2) &= \frac{1}{{16}} \\ I(3) &= \frac{1}{4} - \frac{{\sqrt 2 }}{{24}} - \frac{{\sqrt 6 }}{{18}} \\ I(\frac{1}{2}) &= \frac{1}{{4\pi }} \\ I(\frac{1}{3}) &= 1 - \frac{{3\sqrt 6 }}{8} \\ I(\frac{2}{3}) &= \frac{1}{3} - \frac{{3\sqrt 3 }}{{16}} + \frac{{\sqrt 3 }}{{8\pi }} \end{aligned}$$

In general, $I(r)$ for rational $r$ is algebraic over $\mathbb{Q}(\pi)$.

---

I just realized the question has been answered here, but that answer is not quite complete.