Prove that $\int_{-\infty}^\infty\frac{e^{-tx^2}}{\cosh\pi x}\,dx\ge\frac{e^{t/2}}{t+1}$ for all $t\in[0,1]$ This question is taken directly from Showing that $z^2 e^{-z^2/2} \int \frac{\phi^2(x)}{\cosh(xz)} \, dx \geq \frac{1}{2\sqrt{\pi}} \frac{z^2}{z^2+1}$ which unfortunately turned out to be untrue for all $z\ge0$.
By plotting the functions, it appears that the inequality is true for $0\le z\le1.$ That is, conjecturally, $$\int_{-\infty}^\infty\frac{e^{-u^2}}{\cosh zu}\,du\ge\frac{e^{z^2/2}\sqrt\pi}{z^2+1},\quad\forall z\in[0,1].$$
To simplify this, we can invoke the elegant identity $$\int_0^\infty\frac{e^{-u^2}}{\cosh\alpha u}\,du=\frac{\sqrt\pi}\alpha\int_0^\infty\frac{e^{-u^2}}{\cosh(\pi u/\alpha)}\,du$$ to obtain the equivalent $$\int_{-\infty}^\infty\frac{e^{-u^2}}{\cosh(\pi u/z)}\,du\ge\frac{ze^{z^2/2}}{z^2+1}.$$
Note the identity can be used as the integrand is an even function.
Substituting $x=u/z$ and $t=z^2$ yields $$\int_{-\infty}^\infty\frac{e^{-tx^2}}{\cosh\pi x}\,dx\ge\frac{e^{t/2}}{t+1}.\tag1$$ Can $(1)$ be proven analytically for all $t\in[0,1]$?
 A: Let $I(t)=\int_{\mathbb{R}} e^{-tx^2}\textrm{sech} (\pi x)\ dx$, $J(t)=\log I(t)$ be the functions defined in Jack D'Aurizio's solution.  In his solution, the main idea was that the graph of convex functions lies above any tangent lines.
Note that $I'(t)=\int_{\mathbb{R}} -x^2 e^{-tx^2} \textrm{sech} (\pi x) \ dx$ and $J'(t)=\frac{I'(t)}{I(t)}$. We have
$$
\begin{align}
J''(t)=\left( \frac{I'(t)}{I(t)}\right)'&=\frac{I''(t) I(t) - (I'(t))^2}{I(t)^2}\\ &= \frac{\iint_{\mathbb{R}^2} \left(\frac{x^4+y^4}2-x^2y^2\right)e^{-t(x^2+y^2)} \textrm{sech}(\pi x)\textrm{sech}(\pi y) dA}{ \iint_{\mathbb{R}^2}e^{-t(x^2+y^2)} \textrm{sech}(\pi x)\textrm{sech}(\pi y) dA}\\
&=\frac{\frac12\iint_{\mathbb{R}^2} (x^2-y^2)^2e^{-t(x^2+y^2)} \textrm{sech}(\pi x)\textrm{sech}(\pi y) dA}{ \iint_{\mathbb{R}^2}e^{-t(x^2+y^2)} \textrm{sech}(\pi x)\textrm{sech}(\pi y) dA}>0
\end{align}
$$
Thus, $J$ is convex.
Using convexity of $I$ and the tangent line at $t=0$, we have by $I'(0)=-1/4$,
$$
I(t)\geq 1-\frac t4. 
$$
We use this for $t\in [0,1/2]$. Then by $1-\frac t4 \geq \frac{e^{t/2}}{1+t}$ on $[0, 1/2]$, we have (1) on this range.
Using convexity of $J$ and the tangent line at $t=1$, we have by
$J'(1)=\frac{I'(1)}{I(1)}$ and $\frac{|I'(1)|}{I(1)}>\frac18$, $I(1)>\frac56$  (numerical results, need verification),
$$
J(t)\geq J'(1)(t-1) + J(1) \geq \frac18 (1-t) + \log(\frac56).
$$
Then we have
$$
I(t)\geq \frac56\exp\left( \frac18 (1-t)\right).
$$
We use this for $t\in [1/2, 1]$. Then by $\frac56\exp\left( \frac18 (1-t)\right)\geq \frac{e^{t/2}}{1+t}$, we have (1) on this range.
A possible route to prove $I(t)\geq \exp\left( -\frac t4 + \frac{t^2}{16}\right)$ is as follows.
If we prove that $J'''(t)<0$ (I could not prove this), then
the $J'$ is concave. With a help of numerical result
$$
J'(\frac34) >  - \frac 14 + \frac18 \cdot \frac34, 
$$
we would have for $t\in [0,\frac34]$,
$$
J'(t)\geq -\frac14 + \frac18 t
$$
Then for $t\in [0, \frac34]$, we have
$$
I(t)\geq \exp\left(-\frac t4 + \frac{t^2}{16}\right).$$
For $t\in [\frac34, 1]$, the previous bound
$$
I(t) \geq \frac56\exp\left( \frac18 (1-t)\right)
$$
is stronger than $\exp\left(-\frac t4 + \frac{t^2}{16}\right)$.
A: You may exploit the fact that
$$ I(t) = 2\int_{0}^{+\infty}\frac{e^{-t x^2}}{\cosh(\pi x)}\,dx $$
like any moment, is a function with a convex logarithm. It follows that the graph of $J(t)=\log I(t)$ over $[0,1]$ lies above any tangent line. We have $J(0)=0$ and
$$ J'(0)=\frac{I'(0)}{I(0)}=I'(0)=2\int_{0}^{+\infty}\frac{-x^2\,dx}{\cosh(\pi x)} = -\frac{1}{4},$$
so $J(t)\geq -\frac{t}{4}$ and $I(t)\geq \exp(-t/4)$ over $[0,1]$. By exploiting the log-convexity properties of $-I'(t)$ and $I''(t)$ and numerical approximations at $t=1$ the inequality can be improved up to $J(t)\geq -\frac{t}{4}+\frac{t^2}{16}$, so
$$ I(t) \geq \exp\left(-\frac{t}{4}+\frac{t^2}{16}\right) $$
which is sharper than $I(t)\geq \frac{e^{t/2}}{1+t}$.
A: Too long for a comment :
We have the obvious inequalities for $x\geq 0$ and $t\in[0,0.25]$ :
$$2\int_{0}^{\infty\ }\frac{e^{-tx^2}}{\cosh\left(\pi x\right)}dx\geq 2\int_{0}^{\infty\ }\frac{e^{-tx}}{\cosh\left(\pi x\right)}dx\geq 2\int_{0}^{\infty\ }\frac{1-xt}{\cosh\left(\pi x\right)}dx$$
So we need to show for $x\geq 0$ and $t\in[0,0.25]$:
$$\frac{e^{\frac{t}{2}}}{t+1}\leq 2\int_{0}^{\infty\ }\frac{1-xt}{\cosh\left(\pi x\right)}dx$$
We have :
$$2\int_{0}^{\infty\ }\frac{1}{\cosh\left(\pi x\right)}dx=1$$
And :
$$2\int_{0}^{\infty\ }\frac{-xt}{\cosh\left(\pi x\right)}dx=-2tC/\pi^2$$
Where C is the Catalan's constant .
So we need to show :
$$\frac{e^{\frac{t}{2}}}{t+1}\leq 1-2tC/\pi^2$$
Wich is easier and true .
Edit we have numerically the inequality for $t\in[0.25,0.75]$ :
$$2\left(\int_{0}^{0.6}-\left(\frac{-1+x^{2}t-\left(3-\left((1-c)+2+ac(2-c)-ac(1-c)\ln a\right)\right)}{\cosh\left(\pi\cdot x\right)}\right)dx+\int_{0.6}^{1}\frac{\left(1-x^{2}t\right)}{\cosh\left(\pi\cdot x\right)}dx\right)-\frac{e^{\frac{t}{2}}}{1+t}> 0$$
Where $c=x^{2}t,a=e^{-1}$
In fact I have used this link lemma 5.1
On the other hand we have the inequalities for $x\in[0,0.6]$ and $t\in[0.25,0.75]$ :
$$e^{-tx^{2}}-1+x^{2}t-\left(3-\left((1-c)+2+ac(2-c)-ac(1-c)\ln a\right)\right)\geq 0$$
And :
$$e^{-tx^2}\geq 1-x^2t$$
Last edit :
It seems we have the inequality for $x\in[0,1.252+5\left(t-1\right)]$ and $t\in[0.95,1]$:
$$-\left(2-\left(2(1-c)+a^{b}c(2-c)-ac(1-c)\ln a\right)-x^{4}t-1\right)\leq e^{-tx^2}$$
Where $b=\frac{\pi}{e}$ and $a=e^{-1}$ , $c=x^{2}t$
Next it seems we have the inequality for $x\in[0,1.252+5\left(t-1\right)]$ and $t\in[0.95,1]$:
$$2\int_{0}^{1.252+5\left(t-1\right)}\frac{-\left(2-\left(2(1-c)+a^{b}c(2-c)-ac(1-c)\ln a\right)-x^{4}t-1\right)}{\cosh\left(\pi x\right)}dx+2\int_{1.252+5\left(t-1\right)}^{\infty}\frac{1}{\cosh\left(\pi x\right)^{2}}dx>\frac{e^{\frac{t}{2}}}{t+1}$$
A: Some thoughts. Let $\sigma^2={1 \over 2 t}$. Then
$$
\int_{-\infty}^\infty\frac{e^{-tx^2}}{\cosh\pi x}\,dx
=
\mathbf{E} \left ( {\sqrt{2 \pi} \sigma \over \cosh \pi X}\right ),
$$
where $X$ is normally distributed with zero mean and variance $\sigma^2$. Since $\cosh x \le e^{x^2 \over 2}$ then $\mathbf{E} \left ( {\sqrt{2 \pi} \sigma \over \cosh \pi X}\right ) \ge \mathbf{E} \left ( {\sqrt{2 \pi} \sigma e^{-{\pi^2 X^2 \over 2 }}}\right )$. Now Jensen's inequality can be used.
A: Attempts:
We have the $\operatorname{sech}$ distribution, so the moments of the function $\operatorname{sech} \pi x$ are known and can be expressed in terms of the Euler functions. We have the formula:
$$\int_{-\infty}^{\infty} \frac{e^{-t x}}{\cosh{\pi x}} dx= \sec \frac{t}{2}$$
for $|\operatorname{Re} t|< \pi$
Now, to estimate the integrals
$$\int_{-\infty}^{\infty} \frac{e^{-t x^2}}{\cosh \pi x}dx$$
we'll use the Gaussian quadrature of order $3$. The orhogonal polynomials for the distrubution $\frac{1}{\cosh\pi x}$ are $1$, $x$, $x^2 - \frac{1}{4}$, $x^3 - \frac{5}{4}x$, $\ldots$. The Gaussian quadrature of order $3$ (exact for polynomials of degree $\le 5$) is
$$\int_{-\infty}^{\infty} \frac{f(x)}{\cosh \pi x} \simeq  \frac{1}{10} f( -\frac{\sqrt{5}}{2})+ \frac{4}{5} f(0) + \frac{1}{10}f( \frac{\sqrt{5}}{2}) $$
For $f(x) = e^{-t x^2}$ we get
$$\int_{-\infty}^{\infty} \frac{e^{-t x^2}}{\cosh \pi x} \simeq \frac{1}{5} e^{-\frac{5}{4} t} + \frac{4}{5}$$
It turns out that the RHS in the above approximation is larger for all $t>0$. To get a lower estimate for $t\in [0,1]$, substitute RHS with
$$\frac{1}{4.5} e^{-\frac{4.5}{3.5} t} + \frac{3.5}{4.5}$$
This is larger than $\frac{e^{\frac{t}{2}}}{1+t}$ for $t\in [0,1]$.
This lower estimate is valid for $t\in [0,1]$. A lower estimate that works for all $t>0$ is given by another Gauss quadrature, $e^{-\frac{t}{4}}$.
A: I use a systematic approach.
$$f(t) = \int_{0}^{\infty}\frac{e^{-t x^2}}{\cosh(\pi x)}\,dx$$
Apply the Laplace transform
$$F(s)= \int_{0}^{\infty}f(t)e^{-ts}\,dt$$
or
$$F(s) = \int_{0}^{\infty}\frac{dx}{(s+x^2)\cosh(\pi x)}=\frac{2}{\sqrt{s}} \int_{0}^{1}\frac{x^{2\sqrt{s}}}{1+x^2}$$
The last integral is due to Hardy
Now,  taylor-expand the integrand in the last integral and integrate term by term to get
$$F(s)= \sum_{k=0}^{\infty}\frac{(-1)^k}{\sqrt{s}(\sqrt{s}+k+\frac{1}{2})}$$
To inverse the received expression we use a table of Laplace transform pairs.
Result
$$f(t)= \sum_{k=0}^{\infty}(-1)^k e^{t(k+\frac{1}{2})^2}\operatorname{erfc\left [\sqrt{t}(k+\frac{1}{2})  \right ]}$$
where $\operatorname{erfc(x)}$ is the Complementary Error Function.
From this result, for large $t$,  the asymptotic behavior of $f(t)$ seems to be
$$\frac{2}{\sqrt{\pi t}}$$
But for moderate values of $t$ as required i am not sure how to get suggested bounds.
But in any case, since the series is alternating, the error should be less than the absolute value of the first omitted term.
