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?


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.

  • $\begingroup$ It is unlikely if $c$ stops at $0$. This already gives an integrand with complimentary error function $\endgroup$ Feb 20, 2021 at 22:54
  • $\begingroup$ @user12588 This is a gentle reminder to consider accepting an answer if your question has been resolved. $\endgroup$
    – Sal
    Feb 27, 2021 at 13:38

2 Answers 2


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 $1$:

enter image description here

  • 1
    $\begingroup$ The final result may be simplified further by re-substituting $ K_0 (p) \sim \sqrt {\frac{\pi }{{2p}}} e^{ - p} $ for large $p$. $\endgroup$
    – Gary
    Mar 6, 2021 at 12:07
  • 1
    $\begingroup$ @Gary Good idea! I also caught a typo in my last expression. $\endgroup$
    – Sal
    Mar 6, 2021 at 15:53

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