Upper and lower bounds for $f(t) = \int_{-\infty}^\infty \frac{t^2 e^{-t^2/2}}{\cosh(tz)} \phi(z) \, dz$

Question

Consider the function $$f(t) = \int_{-\infty}^\infty \frac{t^2 e^{-t^2/2}}{\cosh(tz)} \phi(z) \, dz.$$ Above $\phi(z)$ is the Gaussian probability density function.

I am interested in sharp upper and lower bounds for $f$.

I started by using the Taylor series for $\log \cosh (w)$, from which we can obtain the bound $$ e^{w^2/2 - w^4/12} \leq \cosh(w) \leq e^{w^2/2}, \qquad\mbox{(1)} $$ for all $w$. This gives us a lower bound $$ f(t) \geq e^{-t^2/2} \frac{t^2}{\sqrt{1 + t^2}}. $$ Unfortunately, I do not know how to evaluate the integral corresponding to the implied upper bound from (1).

One can also use $\cosh(w) \geq 1 + w^2/2$, and that gives us the relation $$ \frac{1}{\sqrt{1 + t^2}} \leq \frac{f(t)}{t^2 e^{-t^2/2}} \leq \sqrt{\pi} e^{1/t^2} \frac{1}{|t|} \mathrm{Erfc}\Big(\frac{1}{|t|}\Big) $$ Simplification of these bound seems to yield $$ c \frac{t^2 e^{-t^2/2}}{|t| + 1} \leq f(t) \leq C \frac{t^2 e^{-t^2/2}}{|t| + 1} $$ for all $t$. Above $c, C > 0$ are two constants.

Is there a better/simpler bound?

Answer 1

Assume that $t\ge 0$. We have \begin{align*} \int_{ - \infty }^{ + \infty } {{\rm e}^{ - \frac{{x^2 }}{2}} \operatorname{sech} (tx)\,{\rm d}x} & = 2\int_0^{ + \infty } {{\rm e}^{ - \frac{{x^2 }}{2}} \frac{2}{{{\rm e}^{tx} + {\rm e}^{ - tx} }}{\rm d}x} \\ & \le 4\int_0^{ + \infty } {{\rm e}^{ - \frac{{x^2 }}{2} - tx} {\rm d}x} = 4\sqrt {\frac{\pi }{2}} {\rm e}^{\frac{{t^2 }}{2}} \operatorname{erfc}\! \left( {\frac{t}{{\sqrt 2 }}} \right) = 4\sqrt 2\, \mathsf{M}\!\left( {\frac{t}{{\sqrt 2 }}} \right) \\ & \le \frac{8}{{t + \sqrt {t^2 + 8/\pi } }} \le 2\sqrt {2\pi } \frac{1}{{\sqrt {1 + t^2 } }}, \end{align*} where $\mathsf{M}$ is Mills' ratio (cf. $(7.8.2)$). Consequently, using your result, $$ \frac{{t^2 }}{{\sqrt {1 + t^2 } }}{\rm e}^{ - \frac{{t^2 }}{2}} \le f(t) \le 2\frac{{t^2 }}{{\sqrt {1 + t^2 } }}{\rm e}^{ - \frac{{t^2 }}{2}} $$ for all $t\ge 0$, and, by the evenness of $f$, for all real $t$. I believe that the constant $2$ in the upper bound can be improved to $\sqrt{\pi/2}$ (which would be the best possible). The constant factor $1$ in the lower bound is sharp (consider $\lim_{t\to 0}f(t)$).

I also determined the large-$t$ asymptotic expansion of $f$: $$ f(t) \sim \sqrt {\frac{\pi }{2}} |t|{\rm e}^{ - \frac{{t^2 }}{2}} \sum\limits_{m = 0}^\infty {\frac{{E_{2m} }}{{m!}}\left( {\frac{\pi }{{2\sqrt{2}\,t}}} \right)^{2m} }, $$ where the $E_m$ are the Euler numbers. It can be shown that the error comitted by truncating this expansion at any finite number of terms is less in absolute value than the magnitude of the first omitted term and has the same sign as that term for all $t\ne 0$. This will lead to a sequence of upper and lower bounds.

Answer 2

I'll assume by $\phi$ you mean $$\phi(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$$ Your function is $$f(s)=\frac{s^2\exp(-s^2/2)}{\sqrt{2\pi}}\underbrace{\int_{-\infty}^\infty \exp(-x^2/2)\operatorname{sech}(sx)\mathrm dx}_{:=g(s)} $$

We now need to replace that pesky $\operatorname{sech}$ with something simpler so we can come up with good bounds.

I'll leave it to you as an exercise to show (or you can just check) that, for $s,x\in\mathbb R$, that $$\exp\left(-\frac{1}{2}s^2x^2\right)\leq \operatorname{sech}(sx)\leq\frac{1}{1+\frac{s^2x^2}{2}}$$

Hence we can make bounds $$\int_{-\infty}^\infty \exp\left(-\frac{1+s^2}{2}x^2\right)\mathrm dx\leq g(s)\leq \int_{-\infty}^\infty \frac{\exp(-x^2/2)}{1+\frac{s^2x^2}{2}}\mathrm dx$$

The left hand integral is of course very easy to evaluate. It is $\sqrt{\frac{2\pi}{1+s^2}}$. As for the right hand integral, we can transform it as follows. First write it as an integral on $(0,\infty)$ and let $\xi =sx/\sqrt{2}$. (Note that WLOG we can let $s>0$ - it is even WRT $s$.) $$\int_{-\infty}^\infty \frac{\exp(-x^2/2)}{1+\frac{s^2x^2}{2}}\mathrm dx=2\int_{0}^\infty \frac{\exp(-x^2/2)}{1+\frac{s^2x^2}{2}}\mathrm dx \\=2\frac{\sqrt{2}}{s}\int_{0}^\infty \frac{\exp\left(-\frac{1}{s^2}\xi^2\right)}{1+\xi^2}\mathrm d\xi$$

Thankfully, this is a well known integral representation of the complementary error function, $$\frac{\pi}{2}\mathrm e^{z^2}\operatorname{erfc}(z)=\int_0^\infty \frac{\exp(-z^2t^2)}{1+t^2}\mathrm dt$$ Where of course $\operatorname{erfc}(z):=1-\operatorname{erf}(z)$.

So $$2\frac{\sqrt{2}}{s}\int_{0}^\infty \frac{\exp\left(-\frac{1}{s^2}\xi^2\right)}{1+\xi^2}\mathrm d\xi=\frac{\pi\sqrt{2}}{s}\mathrm e^{1/s^2}\operatorname{erfc}(1/s) \\ s>0$$ We can extend this to $s<0$ by taking an absolute value: $$2\frac{\sqrt{2}}{s}\int_{0}^\infty \frac{\exp\left(-\frac{1}{s^2}\xi^2\right)}{1+\xi^2}\mathrm d\xi=\frac{\pi\sqrt{2}}{|s|}\mathrm e^{1/s^2}\operatorname{erfc}(1/|s|) \\ s\in\Bbb R$$

Hence $$\sqrt{\frac{2\pi}{1+s^2}}\leq g(s)\leq \frac{\pi\sqrt{2}}{|s|}\mathrm e^{1/s^2}\operatorname{erfc}(1/|s|)$$

And so finally $$\boxed{\frac{s^2\mathrm e^{-s^2/2}}{\sqrt{1+s^2}}\leq f(s)\leq \sqrt{\pi}~|s|\exp\left(-\frac{s^2}{2}+\frac{1}{s^2}\right)\big(1-\operatorname{erf}(1/|s|)\big)}$$

You can check this numerically. You might notice some issues near zero because Desmos is struggling with the numerical precision.

Answer 3

In the same spirit as @K.defaoite, we can derive close upper bounds for $$g(s)=\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}}\, \text{sech}(s x)\,dx$$ since, using Padé approximants, $$ \text{sech}(t) <\frac {15120-660 t^2+13 t^4}{15120+6900 t^2+313 t^4}=P_4(t)$$ whose error is $\frac{59 t^{10}}{152409600}$.

Making $t=s x$

$$\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}}\, P_4(s x)\,dx$$ write it as $$\frac {13}{313}\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}}\,\frac{(s^2x^2-a)(s^2 x^2-b) } {(s^2x^2-c)(s^2 x^2-d) }\,dx$$ where $(a,b)$ are complex and $(c,d)$ real (fortunately negative).

Using partial fraction decomposition $$\frac{(s^2x^2-a)(s^2 x^2-b) } {(s^2x^2-c)(s^2 x^2-d) }=\frac{a b-a c-b c+c^2}{(c-d) \left(s^2 x^2-c\right)}-\frac{a b-a d-b d+d^2}{(c-d) \left(s^2 x^2-d\right)}+1$$ $$\int_{-\infty}^{+\infty}\frac{e^{-\frac{x^2}{2}}}{s^2 x^2-k}\,dx=\frac \pi{s^2}\sqrt{-\frac{s^2}{k}}\,\,e^{-\frac{k}{2 s^2}}\,\,\text{erfc}\left(\sqrt{-\frac{k}{2 s^2}}\right)$$

Some numbers

$$\left( \begin{array}{cccc} s & \text{numerical integration} & \text{right bound}\\ 1 & 1.8580923 & 1.8580952 \\ 2 & 1.2705356 & 1.2712260 \\ 3 & 0.9366468 & 0.9396230 \\ 4 & 0.7339053 & 0.7405651 \\ 5 & 0.6005304 & 0.6116231 \\ 6 & 0.5070120 & 0.5227788 \\ 7 & 0.4381425 & 0.4585291 \\ 8 & 0.3854699 & 0.4102545 \\ 9 & 0.3439410 & 0.3728508 \\ 10 & 0.3103977 & 0.3431320 \\ \end{array} \right)$$

This could be still improved since $ \text{sech}(t) < P_8(t)$ wich will lead to the same integrals.

Edit

For a lower bound, we can use $$\frac{e^{\frac{1}{2} \left(\frac{8}{\pi ^2}-1\right) t^2}}{\frac{4 t^2}{\pi ^2}+1} ~<~\text{sech}(t)$$ Have a look at formula $(9)$ here.

This would give $$g(s) ~>~\frac{\pi ^2}{2 \sqrt{s^2}} \,\exp\left(\frac{1}{8} \pi ^2 \left(\frac{1}{s^2}+1\right)-1 \right)\,\text{erfc}\left(\frac{1}{2}\sqrt{ \frac{\left(\pi ^2-8\right) s^2+\pi ^2}{2 s^2}}\right)$$

$$\left( \begin{array}{ccc} s & \text{left bound} & \text{numerical integration}\\ 1 & 1.8557495 & 1.8580923 \\ 2 & 1.2633726 & 1.2705356 \\ 3 & 0.9274845 & 0.9366468 \\ 4 & 0.7246235 & 0.7339053 \\ 5 & 0.5918148 & 0.6005304 \\ 6 & 0.4990303 & 0.5070120 \\ 7 & 0.4308830 & 0.4381425 \\ 8 & 0.3788520 & 0.3854699 \\ 9 & 0.3378910 & 0.3439410 \\ 10 & 0.3048403 & 0.3103977 \\ \end{array} \right)$$