# Integral of $\int_{-\infty}^\infty\frac{e^{-y^2/2} \sinh(cy)^2} {\cosh(Mcy) }dy$

Is it difficult to compute or find a good computable lower bound on the integral \begin{align*} \int_{-\infty}^\infty\frac{e^{-y^2/2} \sinh(cy)^2} {\cosh(Mcy) }dy \end{align*} where $c$ and $M$ are constants.

It's not hard to see that the integral is convergent but how should this be approached?

I feel like this integral has already been looked at?

Notice that

$$I = I(M,c) := \int_{-\infty}^{\infty} \frac{e^{-y^2/2}\sinh^2 (cy)}{\cosh(Mcy)} \, dy = \frac{1}{c} \int_{-\infty}^{\infty} \frac{e^{-y^2/2c^2}\sinh^2 y}{\cosh(My)} \, dy.$$

Then as $c \to \infty$ and $M > 2$,

$$I \sim \frac{1}{c} \int_{-\infty}^{\infty} \frac{\sinh^2 y}{\cosh(My)} \, dy = \frac{\pi}{2Mc}(\sec(\pi/M) - 1).$$

• Could you explain ho you pooled the $c$ out of hyperbolic functions? Change of variable right? – Boby May 15 '15 at 22:54
• Also, the answer to the second integral is it $\sec$ or $\sech$?? – Boby May 15 '15 at 23:35
• @Boby, That's right. It is change of variable. Also, $\sec$ is correct. For example, the integral diverges to $\infty$ as $M \to 2^+$, and this phenomenon is exactly replicated by the answer. – Sangchul Lee May 16 '15 at 2:13
• Could you point me a table where you got that integral? Also, do you think it is possible to give a good lower bound for any $c$? – Boby May 16 '15 at 2:49
• Could you please give more details. For example why does $M>2$ and how did you do the last integral? – Boby May 16 '15 at 21:46