How big is the integral $\int_0^\infty \frac{x\exp(-x^2/4)\cosh(x)}{\sqrt{\cosh(x)-1}} dx$

Question

I can't seem to get Maple to approximate the integral

$$\int_0^\infty \frac{x\exp(-x^2/4)\cosh(x)}{\sqrt{\cosh(x)-1}} dx.$$

Could somebody tell me why?

This integral "should be" well-defined. (My reasons are not mathematical. The book I'm reading suggest that this integral makes sense.) Do note that the denominator of the integrand explodes at $x=0$, but this should not be a problem...

Can we give an upper bound for this integral?

Answer 1

Using the following inequality: $$ \frac{x}{\sqrt{\cosh{x}-1}} = \sqrt{2} \frac{x/2}{\sinh(x/2)} \leqslant \sqrt{2} $$ It is easy to work out the upper bound: $$ \int_0^\infty \frac{x\exp(-x^2/4)\cosh(x)}{\sqrt{\cosh(x)-1}} \mathrm{d}x < \sqrt{2} \int_0^\infty \exp(-x^2/4)\cosh(x) \mathrm{d}x = \sqrt{2 \pi} \mathrm{e} \approx 6.8 $$

Answer 2

With a little bit more work you can use the inequality given by Sasha to get a tighter upper bound. Indeed, $$ \frac{\cosh x}{\sqrt{\cosh x -1}}=\frac{1+2\,\sinh^2(x/2)}{\sqrt{2}\sinh(x/2)}=\frac{1}{\sqrt{2}\sinh(x/2)}+\sqrt{2}\sinh(x/2), $$ which splits the original integral into two terms. The first of them may be estimated using Sasha's inequality: $$ \int_0^{\infty} \frac{x\exp(-x^2/4)}{\sqrt{2}\sinh(x/2)}dx=\sqrt{2}\int_0^{\infty} \frac{(x/2)\exp(-x^2/4)}{\sinh(x/2)}dx<\sqrt{2}\int_0^{\infty} \exp(-x^2/4)dx=\sqrt{2\pi}. $$ The second integral computes exactly: $$ \sqrt{2}\int_0^{\infty} x\exp(-x^2/4)\sinh(x/2) dx=\sqrt{2\pi}\mbox{e}^{1/4}, $$ (see section 3.562 in Gradshteyn & Ryzhik's "Table of Integrals, Series and Products" or simply use Mathematica or Maple). Therefore, we obtain $$ \int_0^{\infty} \frac{x\exp(-x^2/4)\cosh x}{\sqrt{\cosh x -1}}dx<\sqrt{2\pi}(1+\mbox{e}^{1/4})\approx 5.7252. $$