If $f(x) = \sum_{n=0}^\infty a_n x^n$, then $\int_{-\infty}^\infty f(x^2) dx = \pi i a_{-\frac{1}{2}}$

Question

I've noticed a curious relationship between the coefficient $a_n$ for a power series and the integral of the real line. For instance, take $f(x) = e^{-x} = \sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n$. Then, we have $a_n = \frac{(-1)^n}{n!}$. The relationship says that $$\int_{-\infty}^\infty e^{-x^2} dx = \pi i a_{\frac{-1}{2}} = \pi i \frac{1}{\sqrt{-1}\Gamma(1/2)}= \sqrt{\pi}$$ This is not at all interesting, the relationship between the integral over $e^{-x^2}$ and the gamma function is easily obtainable by a change of variables.

Here's another example. We have the power series $\frac{\Gamma'(1+x)}{x\Gamma(1+x)}+\frac{\gamma}{x} = \sum_{k=0}^\infty \zeta(k+2) (-1)^k x^{k}$. So the relationship suggests that $$\int_{-\infty}^\infty \Big( \frac{\Gamma'(1+x^2)}{x\Gamma(1+x^2)}+\frac{\gamma}{x^2} \Big) dx= \pi i a_{\frac{-1}{2}}= \pi \zeta(3/2)$$ This (I don't think) follows directly from a change of variables, but it's easily provable using the residue theorem.

Two other examples that fit this pattern are $$\frac{x}{e^x -1} = \sum_{n=0}^\infty \frac{B_n}{n!} x^n$$ Then, noting that $B_n = -n \zeta(1-n)$ gives $$\int_{-\infty}^\infty \frac{x^2}{e^{x^2}-1} dx = \frac{\sqrt{\pi}}{2} \zeta(3/2)$$ or using $\frac{1}{\cosh(x)} = \sum_{n=0}^\infty \frac{E_n}{n!} x^n$ and the fact that $E_n = -\frac{\cos(\pi n/2)e^{ \pi i n}}{ (2 \pi)^n}\left(\zeta(n+1, \frac{1}{4}) - \zeta(n+1, \frac{3}{4}) \right)$ gives $$\int_{-\infty}^\infty \frac{1}{\cosh(x^2)} dx = \sqrt{\pi}\left( \zeta\left(\frac{1}{2},\frac{1}{4}\right) - \zeta\left(\frac{1}{2}, \frac{3}{4}\right)\right)$$

Question

I'm interested if there are any arguments for why this relationship might hold. Are there conditions we can place on $f(x)$ or $a_n$ so that this holds true in general?

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Note: I think there might be a simple proof if we decide to extend Cauchy's Integral theorem to define $$a_n = \int_C \frac{f(z)}{z^{n+1}} dz,$$ where I think $C$ needs to be a curve that goes around the branch cut.

Answer 1

Let $\phi(z)$ be analytic function defined on the half-plane $$H(\delta)=\{z\in \mathbb{C}: \operatorname{Re}z\ge -\delta\}$$ for $0<\delta<1$. Suppose that, for some $A<\pi$, $\phi$ satisfies the growth condition $$|\phi(x+iy)|\le Ce^{Px+ A|y|}$$ for all $z\in H(\delta)$. Then, for any $0<\operatorname{Re} s< \delta$, Ramanujan's master theorem tells you that if $$f(x)=\sum_{k=0}^\infty \phi(k) (-x)^k,$$ then $$\int_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$ Therefore, $$\int_0^\infty x^{2s-1} f\left(x^2\right)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$ so that, letting $s=1/2$, $$\int_0^\infty f\left(x^2\right)\,dx=\frac{\pi}{2}\phi(-1/2). $$ The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$ as long as the hypotheses of the theorem are satisfied.