0
$\begingroup$

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?

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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 $M > 2$,

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

$\endgroup$
5
  • $\begingroup$ Could you explain ho you pooled the $c$ out of hyperbolic functions? Change of variable right? $\endgroup$
    – Boby
    May 15, 2015 at 22:54
  • $\begingroup$ Also, the answer to the second integral is it $\sec$ or $\sech$?? $\endgroup$
    – Boby
    May 15, 2015 at 23:35
  • $\begingroup$ @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. $\endgroup$
    – Sangchul Lee
    May 16, 2015 at 2:13
  • $\begingroup$ 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 $c$? $\endgroup$
    – Boby
    May 16, 2015 at 2:49
  • $\begingroup$ Could you please give more details. For example why does $M>2$ and how did you do the last integral? $\endgroup$
    – Boby
    May 16, 2015 at 21:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .