# Integral involving exponentials of cosh functions

I recently encountered the following integral

$$I_p\equiv \int\limits_0^{+\infty}\,\int\limits_{-\infty}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}x\,\mathrm{d}c\,,$$

with $$p>0$$. Can $$I_p$$ be calculated in terms of elementary functions?

Update:

Using the following integral representation for the Bessel function

$$\int\limits_0^{+\infty}e^{-a\cosh x}\;\mathrm{d}x=K_0(a)\,,$$

one can bring $$I_p$$ to the following form:

$$I_p=2 \int\limits_0^{+\infty}e^{-c\,p}\;K_0\left[\left(c^2+1\right) p\right]\;\mathrm{d}c\,.$$

Which shows that $$I_p$$ can be seen as a Laplace transform. However, I haven't managed to do the last integral either.

• It is unlikely if $c$ stops at $0$. This already gives an integrand with complimentary error function Feb 20 at 22:54
• @user12588 This is a gentle reminder to consider accepting an answer if your question has been resolved.
– Sal
Feb 27 at 13:38

By Fubinis theorem: $$I_p = \int\limits_0^{+\infty}\,\int\limits_{-\infty}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}x\,\mathrm{d}c =\int\limits_{-\infty}^{+\infty}\,\int\limits_{0}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}c\,\mathrm{d}x.$$ We can solve the integral with respect to $$c$$ analytically: $$\int\limits_{0}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}c = -\dfrac{\sqrt{{\pi}}\mathrm{e}^{\frac{p}{4\cosh\left(x\right)}-p\cosh\left(x\right)}\left(\operatorname{erf}\left(\frac{\sqrt{p}}{2\sqrt{\cosh\left(x\right)}}\right)-1\right)}{2\sqrt{p\cosh\left(x\right)}}$$ I am afraid that we can't solve the second integral analytically.

Given the mess that results in doing just one of the integrals, I think it would be shocking if your integral had a closed form in elementary functions. We will find the large $$p$$ behavior. Let

$$I(p)=\int\limits_0^\infty dx \ e^{-px} K_0\left[p(x^2+1)\right]$$

Using the asymptotic expansion for $$K_0$$

$$K_0(z)\sim\sqrt{\frac{\pi}{2z}}e^{-z}\sum\limits_{n=0}^\infty c_n z^{-n} \ \ , \ z \to \infty$$

We will not need the form of the $$c_n$$ here. When we substitute $$z=p(x^2+1)$$, we will get a Gaussian $$e^{-p(x^2+1)}$$; we will use (essentially) Laplace's method to do the integral for large $$p$$. When we do, all the factors of $$(x^2+1)$$ outside the exponent will be evaluated at the maxima $$x=0$$. We will be left with

$$\sqrt{\frac{\pi}{2p}}e^{-p(x^2+1)} \sum\limits_{n=0}^\infty c_np^{-n}$$

But this simplifies because the sum is known

$$K_0(p)\sim\sqrt{\frac{\pi}{2p}}e^{-p}\sum\limits_{n=0}^\infty c_np^{-n} \ \ , \ \ p \to \infty$$

So we get

$$I(p)\sim \int\limits_0^\infty dx \ e^{-px} e^{p-p(x^2+1)} K_0(p) \ \ , \ \ p \to \infty$$

The integral is an error function

$$I(p) \sim \frac{1}{2}\sqrt{\frac{\pi}{p}}e^{p/4}\operatorname{erfc}\left(\frac{\sqrt{p}}{2} \right) K_0(p) \ \ , \ \ p \to \infty$$

We can simplify further (thanks @Gary). Expand $$K_0$$ and $$\operatorname{erfc}$$ for large arguments, yielding

$$I(p) \sim \sqrt{\frac{\pi}{2}} p^{-3/2}e^{-p} \quad , \quad p \to \infty$$

Here is a plot of the approximations versus numeric result for 'large' values of $$p$$ less than $$1$$:

• The final result may be simplified further by re-substituting $K_0 (p) \sim \sqrt {\frac{\pi }{{2p}}} e^{ - p}$ for large $p$.
– Gary
Mar 6 at 12:07
• @Gary Good idea! I also caught a typo in my last expression.
– Sal
Mar 6 at 15:53