Is there a way to find the antiderivative of this infinite product?

Question

There probably isn't a clean answer for what I'm asking for but I'm curious to see if there is any way to find an antiderivative for this function: $$f(x)=\prod_{n=1}^{\infty} \frac{1}{n^{2}+x^{2}}$$ Using basic contour integration over a semicircular path and residue theorem I got: $$ \int_{-\infty}^{\infty}\left (\prod_{n=1}^{\infty} \frac{1}{n^{2}+x^{2}} \right )dx=0$$
Since the function slowly resembles a line $y=0$ as more terms are being multiplied in the infinite product, its antiderivative should also be $0$ which would make it a whole lot less exciting, so I'm curious to see if there is any other way to express the antiderivative of this function.

Answer 1

Since $f(x)$ is a product of positive terms, we know $\forall x \in \Bbb R~(f(x) \geq 0)$. Thus, if you're correct that $\int_{- \infty}^{\infty} f(x)~dx=0$, it must be the case that $f(x)=0$, which means that its antiderivative is a constant function.

Answer 2

$$\int_{-\infty}^{+\infty}\prod_{r=1}^{a} \frac{1}{r^{2}+x^{2}} \mathrm dx=\frac{\pi}{(a)!(a-1)!(2a-1)}$$

Well it seems the question has been answered but to add on, $$f(x)=\prod_{n=1}^{\infty} \frac{1}{n^{2}+x^{2}}$$ Also, $$\Gamma(ix-n)\Gamma(-ix-n)=\frac\pi{x\sinh(\pi x)}\prod_{n=1}^n\frac1{n^2+x^2}$$

$$f(x)=\lim_{n \to \infty }\Gamma(ix-n)\Gamma(-ix-n) \frac{x\sinh(\pi x)}{\pi}$$ $$f(x)=\frac{x\sinh(\pi x)}{\pi} \lim_{n \to \infty }\Gamma(ix-n)\Gamma(-ix-n)$$ The limit indeed approaches zero implying $$f(x)=0$$

Graph when $n = 10$

Graph when $n=30$

Graph when $n=90$

It always seems to have that peak.

EDIT: Just something interesting: Instead of $f(x)$ a better function would be: $$g(x)=(a!)^{2}\prod_{r=1}^{a}\frac{1}{r^2+x^2}$$ This modification adjusts the function such that $g(0)=1$ always

The integral becomes: $$\int_{-\infty}^{+\infty}(a!)^{2}\prod_{r=1}^{a}\frac{1}{r^2+x^2} \mathrm dx$$ $$\int_{-\infty}^{+\infty}g(x)\mathrm dx = \frac{a}{2a-1}\pi$$ $$\lim_{a \to \infty} \int_{-\infty}^{+\infty}g(x)\mathrm dx =\lim_{a \to \infty} \frac{a}{2a-1}\pi$$ $$=\lim_{a \to \infty} \frac{1}{2-\frac{1}{a}}\pi$$ $$=\frac{\pi}{2}$$

EDIT: Using this we can find the formula for the integration of partial product: $$\int_{-\infty}^{+\infty}\prod_{r=1}^{a} \frac{1}{r^{2}+x^{2}} \mathrm dx=\frac{\pi}{(a)!(a-1)!(2a-1)}$$

Answer 3

What could be intersting is to look at $$I_m=\int_{0}^{\infty}\left (\prod_{n=1}^{m} \frac{1}{n^{2}+x^{2}} \right )dx$$

Concerning the antiderivative, using partial fraction decomposition, we end with a sum of arctangents which does not make any problem. Using Pochhammer symbols $$I_m=\frac{\pi }{2 (2 m-1) (1)_{m-1} (2)_{m-1}}$$ which gives the recurrence relation $$I_{m+1}=\frac{(2 m-1)}{m (m+1) (2 m+1)} I_m\qquad \text{with} \qquad I_1=\frac \pi 2$$