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Let $k \ge 0$ be an integer. Consider the function

$$ F_k(x, y) = \int_{-\infty}^\infty (x + i w)^k e^{-\pi(x^2 + w^2)} e^{-2\pi i w y} \mathrm{d}w.$$

These functions are all of the form $e^{-\pi(x^2 + y^2)} \times $ (polynomial in $x, y$). Is it true that $$\int_0^\infty F_k(x, \tfrac{1}{x})\, x^{k-2} \mathrm{d}x = 2^{k-1} e^{-2\pi} ?$$


Here's how far I got. Playing with contour integrals and formulae for the Gamma function, I managed to show that $$F_k(x, y) = e^{-\pi(x^2 + y^2)} \sum_{j = 0}^{\lfloor k/2\rfloor} \binom{k}{2j}(-1)^j \frac{(2j)!}{2^{2j} j! \pi^j} (x + y)^{k-2j},$$ and we have $$ \int_0^\infty x^\alpha e^{-\pi(x^2 + 1/x^2)}\ dx = K_{(1 + \alpha)/2}(2\pi)$$ where $K_\nu$ is the "modified Bessel function of the second kind" of order $\nu$. In my formulae $\alpha$ will be an even integer; so my integral is some linear combination of values of the Bessel function at half-integer orders. However, the coefficients are rather messy expressions in $\pi$; for instance when $k = 2$ we get $$ K_{-1/2}(2\pi) + (2 - \tfrac{1}{2\pi}) K_{1/2}(2\pi) + K_{3/2}(2\pi).$$ Wolfram Alpha confirms that the values are the right ones for the first few values of $k$, but what's making these miraculously simplify down? (The integrals for each individual summand in my formula for $F_k$ doesn't seem to have have a nice form.)

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1 Answer 1

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With the change of integration variables $w=xt$, your $F_k (x,y)$ becomes $$ F_k (x,y) = x^{k + 1} \int_{ - \infty }^{ + \infty } {(1 + {\rm i}t)^k {\rm e}^{ - \pi x^2 (1 + t^2 )} {\rm e}^{ - 2\pi {\rm i}txy} {\rm d}t} . $$ Then, by interchanging the order of integration, \begin{align*} \int_0^{+\infty} F_k\!\left(x, \tfrac{1}{x}\right) x^{k-2} \mathrm{d}x & =\int_{ - \infty }^{ + \infty } {(1 + {\rm i}t)^k {\rm e}^{ - 2\pi {\rm i}t} \int_0^{ + \infty } {x^{2k - 1} {\rm e}^{ - \pi x^2 (1 + t^2 )} {\rm d}x}\, {\rm d}t} \\ & = \frac{1}{{\pi ^k }}\int_{ - \infty }^{ + \infty } {\frac{{(1 + {\rm i}t)^k }}{{(1 + t^2 )^k }}{\rm e}^{ - 2\pi {\rm i}t} \int_0^{ + \infty } {s^{2k-1} {\rm e}^{ - s^2 } {\rm d}s}\, {\rm d}t} \\ & = \frac{1}{{\pi ^k }}\frac{{\Gamma (k)}}{2}\int_{ - \infty }^{ + \infty } {\frac{{(1 + {\rm i}t)^k }}{{(1 + t^2 )^k }}{\rm e}^{ - 2\pi {\rm i}t} {\rm d}t} \\ & = \frac{1}{{\pi ^k }}\frac{{\Gamma (k)}}{2}\int_{ - \infty }^{ + \infty } {\frac{{{\rm e}^{ - 2\pi {\rm i}t} }}{{(1 - {\rm i}t)^k }}{\rm d}t} \\ & = 2^{k - 1} {\rm e}^{ - 2\pi } \frac{{\Gamma (k)}}{{2\pi {\rm i}}}\int_{1 - {\rm i}\infty }^{1 + {\rm i}\infty } {\frac{{{\rm e}^u }}{{u^k }}{\rm d}u} \\ & = 2^{k - 1} {\rm e}^{ - 2\pi } . \end{align*} Here I made the change of variables from $x$ to $s$ via $s=\sqrt{\pi}x\sqrt{1+t^2}$, then from $t$ to $u$ via $u=2\pi(1-\mathrm{i}t)$. In the last step, I used Hankel's loop integral for the gamma function. Alternatively we can push the contour of integration accross the pole at $u=0$ to infinity and use the residue theorem.

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  • $\begingroup$ Very nice! The last step looks rather reminiscent of a Mellin inversion formula, I will have to think if that can be made precise. $\endgroup$ Feb 10, 2023 at 2:59
  • $\begingroup$ @DavidLoeffler You can deform the contour of integration into a Hankel loop contour around the negative real axis. $\endgroup$
    – Gary
    Feb 10, 2023 at 3:14

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