An integral involving a hyperbolic secant function and a Gaussian

Question

I am interested in the following integral $$ f(\tau, n) = \int_{-\infty}^\infty \frac{e^{-\tau x^2 / \pi}}{\cosh(x)^n} dx\ . $$ (It has some physics applications to localization of path integrals of 3d supersymmetric gauge theories on a three sphere.) The parameter $\tau$ should have positive real part, and I was interested in the case where $n$ was a positive integer. I was wondering if anyone knows how to express this integral in terms of known special functions, perhaps using contour integral methods, or in fact anything that works.

I have noticed a couple of interesting features of this integral. For example, using the fact that the Fourier transform of $1/\cosh(x)$ is again a hyperbolic secant, one can rewrite this integral as $$ f(\tau,n) = \frac{\pi}{\sqrt{\tau}} \int \frac{ e^{- (\sum_i x_i)^2 /\pi \tau}}{\prod_{i=1}^n \pi \cosh(x_i)} d^n x \ . $$ In the special case $n=1$, this relation looks a bit like the action of $\tau \to -1/\tau$ of $SL(2,{\mathbb Z})$ on a possibly modular function, $$ f(1/\tau,1) = \sqrt{\tau} \, f(\tau,1) \ . $$ However, I don't seem to find an analog result for $\tau \to \tau +1$.

One can evaluate the integral at large $\tau$ and/or $n$ using saddle point integration, yielding $$ f(\tau, n) = \frac{\pi}{\sqrt{\tau + \frac{\pi n}{2}}} \ . $$ Finally, one can evaluate the integral exactly in the limit $\tau = 0$.

But it feels like one ought to be able to do a little more. Suggestions?

Answer 1

Mathematica 11.3 gives approximation solution for the limit $\tau =0$

AsymptoticIntegrate[Exp[-\[Tau]*x^2/Pi]/
Cosh[x]^n, {x, -Infinity, Infinity}, {\[Tau], 0, 5}, 
Assumptions -> {n > 0, n \[Element] Integers, \[Tau] >= 0}]

(* (2^(1 + n) Hypergeometric2F1[n/2, n, (2 + n)/2, -1])/n - (
2^(2 + n) \[Tau] HypergeometricPFQ[{n/2, n/2, n/2, n}, {1 + n/2, 
1 + n/2, 1 + n/2}, -1])/(n^3 \[Pi]) + (
3 2^(3 + n) \[Tau]^2 HypergeometricPFQ[{n/2, n/2, n/2, n/2, n/2, 
n}, {1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2}, -1])/(
n^5 \[Pi]^2) - (1/(n^7 \[Pi]^3))
15 2^(4 + 
n) \[Tau]^3 HypergeometricPFQ[{n/2, n/2, n/2, n/2, n/2, n/2, n/2, 
n}, {1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 
1 + n/2}, -1] + (1/(n^9 \[Pi]^4))
105 2^(5 + 
n) \[Tau]^4 HypergeometricPFQ[{n/2, n/2, n/2, n/2, n/2, n/2, n/2, 
n/2, n/2, n}, {1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 
1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2}, -1] - 
1/(n^11 \[Pi]^5)945 2^(
6 + n) \[Tau]^5 HypergeometricPFQ[{n/2, n/2, n/2, n/2, n/2, n/2, n/
2, n/2, n/2, n/2, n/2, n}, {1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 
1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 1 + n/2, 
1 + n/2}, -1] *)

$$-\frac{945\ 2^{n+6} \tau ^5 \, _{12}F_{11}\left(\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n} {2},n;\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1 ,\frac{n}{2}+1;-1\right)}{\pi ^5 n^{11}}+\frac{105\ 2^{n+5} \tau ^4 \, _{10}F_9\left(\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},n;\frac{n}{2}+1,\frac{n }{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1;-1\right)}{\pi ^4 n^9}-\frac{15\ 2^{n+4} \tau ^3 \, _8F_7\left(\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},n;\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{ n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1;-1\right)}{\pi ^3 n^7}+\frac{3\ 2^{n+3} \tau ^2 \, _6F_5\left(\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},\frac{n}{2},n;\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1;-1 \right)}{\pi ^2 n^5}-\frac{2^{n+2} \tau \, _4F_3\left(\frac{n}{2},\frac{n}{2},\frac{n}{2},n;\frac{n}{2}+1,\frac{n}{2}+1,\frac{n}{2}+1;-1\right)}{\pi n^3}+\frac{2^{n+1} \, _2F_1\left(\frac{n}{2},n;\frac{n+2}{2};-1\right)}{n}$$

and for large $ \tau =\infty$:

AsymptoticIntegrate[Exp[-\[Tau]*x^2/Pi]/
Cosh[x]^n, {x, -Infinity, Infinity}, {\[Tau], Infinity, 10}, 
Assumptions -> {n > 0, n \[Element] Integers, \[Tau] >= 0}]

(* -((31 n \[Pi]^6)/(480 \[Tau]^(11/2))) - (79 n^2 \[Pi]^6)/(
512 \[Tau]^(11/2)) - (273 n^3 \[Pi]^6)/(2048 \[Tau]^(11/2)) - (
105 n^4 \[Pi]^6)/(2048 \[Tau]^(11/2)) - (63 n^5 \[Pi]^6)/(
8192 \[Tau]^(11/2)) + (17 n \[Pi]^5)/(384 \[Tau]^(9/2)) + (
49 n^2 \[Pi]^5)/(512 \[Tau]^(9/2)) + (35 n^3 \[Pi]^5)/(
512 \[Tau]^(9/2)) + (35 n^4 \[Pi]^5)/(2048 \[Tau]^(9/2)) - (
n \[Pi]^4)/(24 \[Tau]^(7/2)) - (5 n^2 \[Pi]^4)/(64 \[Tau]^(7/2)) - (
5 n^3 \[Pi]^4)/(128 \[Tau]^(7/2)) + (n \[Pi]^3)/(16 \[Tau]^(5/2)) + (
3 n^2 \[Pi]^3)/(32 \[Tau]^(5/2)) - (n \[Pi]^2)/(
4 \[Tau]^(3/2)) + \[Pi]/Sqrt[\[Tau]] *)

$$-\frac{63 \pi ^6 n^5}{8192 \tau ^{11/2}}+\frac{35 \pi ^5 n^4}{2048 \tau ^{9/2}}-\frac{105 \pi ^6 n^4}{2048 \tau ^{11/2}}-\frac{5 \pi ^4 n^3}{128 \tau ^{7/2}}+\frac{35 \pi ^5 n^3}{512 \tau ^{9/2}}-\frac{273 \pi ^6 n^3}{2048 \tau ^{11/2}}+\frac{3 \pi ^3 n^2}{32 \tau ^{5/2}}-\frac{5 \pi ^4 n^2}{64 \tau ^{7/2}}+\frac{49 \pi ^5 n^2}{512 \tau ^{9/2}}-\frac{79 \pi ^6 n^2}{512 \tau ^{11/2}}-\frac{\pi ^2 n}{4 \tau ^{3/2}}+\frac{\pi ^3 n}{16 \tau ^{5/2}}-\frac{\pi ^4 n}{24 \tau ^{7/2}}+\frac{17 \pi ^5 n}{384 \tau ^{9/2}}-\frac{31 \pi ^6 n}{480 \tau ^{11/2}}+\frac{\pi }{\sqrt{\tau }}$$

for general values $\tau$ and $n$ can't find closed-form function.