Closed form for $I= \int_{-\infty}^{\infty} e^{-e^x-e^{-x}+x}~ dx$

Question

Is there a closed form for this integral?

$$I= \int_{-\infty}^{\infty} e^{-e^x-e^{-x}+x}~ dx$$

I tried $u=e^x$ but it didn't simplify things much.

I think the closed form is a Bessel function, but I can't verify this directly.

I'd like to see the steps involved to reach the closed form.

Answer 1

You can use the integral representation $$ K_\nu (z) = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\mathrm{e}^{ - z\cosh x + \nu x} \mathrm{d}x} \quad (\Re z>0) $$ of the modified Bessel function to conclude that $I = 2K_1 (2)$.

Answer 2

The substitution actually helps quite a bit. $$I=\int_{-\infty}^\infty \exp(-e^x-e^{-x}+x)\mathrm{d}x$$ Using $u=e^x$, $$I=\int_0^\infty \exp\left(-u-\frac{1}{u}+\ln u\right)\frac{\mathrm{d}u}{u}=\int_0^\infty\exp\left(-u-\frac{1}{u}\right)\mathrm{d}u$$ Using the integral identity I referenced in your previous question this is $$I=2K_{-1}(2)=2K_1(2)$$