# 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 $$t\in[0,1]$$?

• Sort of. You can reduce it to the case $t=1$ and one additional simple numeric inequality, but how are you going to verify those without computations? Commented Jun 10, 2022 at 4:14
• @fedja If your whole proof simple? (I also have a proof assuming that it is not difficult to verify something like $\int_0^\infty \frac{\mathrm{e}^{-x^2-1/2}}{\cosh \pi x}\,\mathrm{d} x > \frac14, \int_0^\infty \frac{x^2\mathrm{e}^{-x^2-1/2}}{\cosh \pi x}\,\mathrm{d} x > \frac{1}{30}$. Commented Jun 10, 2022 at 6:30
• @RiverLi The reduction is rather simple (take the logarithmic derivative of both sides and go back from the inequality at $t=1$), but the case $t=1$ is rather delicate: we have a leeway of just about $0.01$, so I don't know how to handle it properly yet. Commented Jun 10, 2022 at 9:17
• @fedja Perhaps no easy way. So your proof is nice (remains to check $t=1$). Commented Jun 10, 2022 at 13:14
• The discussion between fedja and River Li seems quite close to the full solution (or already done) upon computation of $I(1)$ and $I'(1)$. It seems we can put together results around $t=1$, by fedja and River Li. Around $t=0$, by Jack. Commented Jun 14, 2022 at 1:42

## 8 Answers

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 $$I(t)\geq \frac{e^{t/2}}{1+t}$$.

• Following your solution, I see that $J'(t)\geq -1/4 + t/4$. So, we might have $J(t)\geq -t/4 + t^2/8$. Commented Jun 11, 2022 at 19:50
• But, $J'(1)\geq 0$ at $t=1$ is false. So, I guess it could not be improved to $t^2/8$. I am getting $J''(0)=1/4$. Maybe I got incorrect $J''(0)$? Commented Jun 11, 2022 at 20:53
• Upon closer inspection, it appears that $J'(t)$ is actually concave so the direction of the inequality is the other way round, unfortunately. Commented Jun 11, 2022 at 21:52
• This is really interesting, as your bound $\exp\left(-\frac t4+\frac{t^2}{16}\right)$ graphically works for $0\le t\le1.268\cdots$ only. However, there is nothing currently to suggest why this wouldn't work for $t>1.268\cdots$, so it may be a bit harder than it seems. Graphically, the upper bound of $\exp\left(-\frac t4+\frac{t^2}8\right)$ holds for all $t\ge0$. Commented Jun 11, 2022 at 21:58
• Okay. Firstly, integrating exponential will give possibly nonzero constant term. Secondly, it is still not clear how $t^2/16$ is obtained from repeated integration of the exponential. Commented Jun 14, 2022 at 20:13

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 $$\exp\left(-\frac t4 + \frac{t^2}{16}\right)$$.

• I'm busy for the rest of the month but I'll attempt to prove the numerical bounds at $t=1$. For $I(1)$ it is possible to write the integral as a series involving $\operatorname{erf}$ and truncate until the $5/6$ is reached. Commented Jun 19, 2022 at 14:20

Alternative proof for the original lower bound:

Problem 1: Let $$0 \le t \le 1$$. Prove that $$2\int_0^\infty\frac{\mathrm{e}^{-tx^2}}{\cosh\pi x}\,\mathrm{d}x \ge \frac{\mathrm{e}^{t/2}}{t+1}.$$

Using $$\mathrm{e}^{-u} \ge(1 - \frac{u}{5})^5$$ for all $$u \ge 0$$ (equivalently $$\mathrm{e}^{-u/5} \ge 1 - u/5$$), we have \begin{align*} &2\int_0^\infty\frac{\mathrm{e}^{-tx^2}}{\cosh\pi x}\,\mathrm{d}x\\ \ge\,& 2\int_0^\infty\frac{(1 - \frac{tx^2}{5})^5}{\cosh\pi x}\,\mathrm{d}x\\ =\,& 2\int_0^\infty\frac{1-t{x}^{2}+\frac25\,{t}^{2}{x}^{4}-{\frac {2}{25}}\,{t}^{3}{x}^{6}+{\frac {1}{125}}\,{t}^{4}{x}^{8}-{\frac {1}{3125}}\,{t}^{5}{x}^{10} }{\cosh\pi x}\,\mathrm{d}x\\ =\,& 1 - \frac14\,t + \frac18\,{t}^{2}-{\frac {61}{800}}\,{t}^{3}+{\frac {277}{6400}}\, {t}^{4}-{\frac {50521}{3200000}}\,{t}^{5} \end{align*} where we have used $$\int_0^\infty \frac{1}{\cosh \pi x}\,\mathrm{d} x = \frac12, \quad \int_0^\infty \frac{x^2}{\cosh \pi x}\,\mathrm{d} x = \frac18,$$ $$\int_0^\infty \frac{x^4}{\cosh \pi x}\,\mathrm{d} x = \frac{5}{32}, \quad \int_0^\infty \frac{x^6}{\cosh \pi x}\,\mathrm{d} x = \frac{61}{128},$$ $$\int_0^\infty \frac{x^8}{\cosh \pi x}\,\mathrm{d} x = \frac{1385}{512}, \quad \int_0^\infty \frac{x^{10}}{\cosh \pi x}\,\mathrm{d} x = \frac{50521}{2048}.$$ (See: The integral : $\frac{1}{2}\int_0^\infty x^n \operatorname{sech}(x)\mathrm dx$)

It suffices to prove that $$1 - \frac14\,t + \frac18\,{t}^{2}-{\frac {61}{800}}\,{t}^{3}+{\frac {277}{6400}}\, {t}^{4}-{\frac {50521}{3200000}}\,{t}^{5} \ge \frac{\mathrm{e}^{t/2}}{t+1}$$ or $$\left(1 - \frac14\,t + \frac18\,{t}^{2}-{\frac {61}{800}}\,{t}^{3}+{\frac {277}{6400}}\, {t}^{4}-{\frac {50521}{3200000}}\,{t}^{5}\right)(1 + t) \ge \mathrm{e}^{t/2}.$$ Let $$g(t) := \mathrm{LHS}$$ and $$h(t) := \mathrm{e}^{t/2}$$. It is easy to prove that $$g(t)$$ is concave on $$[0, 1]$$. So $$g(t) - h(t)$$ is concave on $$[0, 1]$$. Also, $$g(0) - h(0) = 0$$ and $$g(1) - h(1) > 0$$. Thus, $$g(t) \ge h(t)$$ on $$[0, 1]$$.

We are done.

Remark 1: The bound is stronger than $$\mathrm{e}^{-t/4 + t^2/17}$$ on $$[0, 1]$$, i.e. for all $$t\in [0, 1]$$, $$1 - \frac14\,t + \frac18\,{t}^{2}-{\frac {61}{800}}\,{t}^{3}+{\frac {277}{6400}}\, {t}^{4}-{\frac {50521}{3200000}}\,{t}^{5} \ge \mathrm{e}^{-t/4 + t^2/17}.$$ However, the bound is weaker than $$\mathrm{e}^{-t/4 + t^2/16}$$ when $$t > 0.887...$$

Remark 2: We can obtain a slightly better bound by using $$\mathrm{e}^{-u} \ge(1 - \frac{u}{7})^7$$.

Sketch of a proof for the lower bound $$\mathrm{e}^{-t/4 + t^2/16}$$

Remarks: In my another answer, I proved a weaker lower bound by using a lower bound for $$\mathrm{e}^{-tx^2}$$. Here, we used another lower bound for $$\mathrm{e}^{-tx^2}$$.

Problem 2: Let $$0 < t \le 1$$. Prove that $$2\int_0^\infty\frac{\mathrm{e}^{-tx^2}}{\cosh\pi x}\,\mathrm{d}x\ge\exp\left(-\frac t4+\frac{t^2}{16}\right).$$

Proof:

Fact 1: For all $$u \ge 0$$, $$\mathrm{e}^{-u} \ge 1 - u + \frac{13}{30}u^2 - \frac{181}{1950} u^3 + \frac{9}{1000}u^4 - \frac{1}{3125}u^5.$$

Using Fact 1, we have \begin{align*} &2\int_0^\infty\frac{\mathrm{e}^{-tx^2}}{\cosh\pi x}\,\mathrm{d}x\\ \ge\,& 2\int_0^\infty\frac{1 - tx^2 + \frac{13}{30}t^2x^4 - \frac{181}{1950} t^3x^6 + \frac{9}{1000}t^4x^8 - \frac{1}{3125}t^5x^{10}}{\cosh\pi x}\,\mathrm{d}x\\ =\,& 1- \frac14\,t+{\frac {13}{96}}\,{t}^{2}-{\frac {11041}{124800}}\,{t}^{3}+{ \frac {2493}{51200}}\,{t}^{4}-{\frac {50521}{3200000}}\,{t}^{5} \end{align*} where we have used $$\int_0^\infty \frac{1}{\cosh \pi x}\,\mathrm{d} x = \frac12, \quad \int_0^\infty \frac{x^2}{\cosh \pi x}\,\mathrm{d} x = \frac18,$$ $$\int_0^\infty \frac{x^4}{\cosh \pi x}\,\mathrm{d} x = \frac{5}{32}, \quad \int_0^\infty \frac{x^6}{\cosh \pi x}\,\mathrm{d} x = \frac{61}{128},$$ $$\int_0^\infty \frac{x^8}{\cosh \pi x}\,\mathrm{d} x = \frac{1385}{512}, \quad \int_0^\infty \frac{x^{10}}{\cosh \pi x}\,\mathrm{d} x = \frac{50521}{2048}.$$ (See: The integral : $\frac{1}{2}\int_0^\infty x^n \operatorname{sech}(x)\mathrm dx$)

It suffices to prove that $$1- \frac14\,t+{\frac {13}{96}}\,{t}^{2}-{\frac {11041}{124800}}\,{t}^{3}+{ \frac {2493}{51200}}\,{t}^{4}-{\frac {50521}{3200000}}\,{t}^{5} \ge \mathrm{e}^{-t/4 + t^2/16}.$$

Let $$f(t) := \mathrm{LHS}$$. Let $$h(t) := 1-\frac14\,t+{\frac {3}{32}}\,{t}^{2}-{\frac {7}{384}}\,{t}^{3}+{\frac {25 }{6144}}\,{t}^{4}-{\frac {27}{40960}}\,{t}^{5}+{\frac {331}{2949120}} \,{t}^{6}.$$ (Note: $$h(t)$$ is the $$6$$-th order Taylor approximation of $$\mathrm{e}^{-t/4 + t^2/16}$$ around $$t = 0$$.)

It suffices to prove that $$f(t) \ge h(t)$$ and $$h(t) \ge \mathrm{e}^{-t/4 + t^2/16}$$ for all $$t\in [0, 1]$$. Omitted.

Proof of Fact 1:

Let $$F(u) := 1 - u + \frac{13}{30}u^2 - \frac{181}{1950} u^3 + \frac{9}{1000}u^4 - \frac{1}{3125}u^5,$$ and $$G(u) := -\frac{u^3 - 12u^2 + 60u - 120}{u^3 + 12u^2 + 60u + 120}.$$ (Note: $$G(u)$$ is $$(3,3)$$-Pade approximant of $$\mathrm{e}^{-u}$$ at $$u = 0$$.)

We have \begin{align*} &G(u) - F(u)\\ =\,& \frac{u^2(312u^6 - 5031u^5 + 3920u^4 + 174440u^3 + 282000u^2 - 4740000u + 7800000)}{975000(u^3 + 12u^2 + 60u + 120)}\\ \ge\,& 0. \end{align*}

It suffices to prove that $$\mathrm{e}^{-u} \ge G(u)$$.

Let $$u_0 = (4 + 4\sqrt 5)^{1/3} - 4(4 + 4\sqrt 5)^{-1/3} + 4.$$ Then $$G(u_0) = 0$$, and $$G(u) > 0$$ on $$[0, u_0)$$, and $$G(u) < 0$$ on $$(u_0, \infty)$$.

Let $$H(u) := -u - \ln G(u)$$. We have, for all $$u \in [0, u_0)$$, $$H'(u) = G(u) \frac{u^6}{(u^3 - 12u^2 + 60u - 120)^2} \ge 0.$$ Also, we have $$H(0) = 0$$. Thus, we have $$H(u) \ge 0$$ on $$(0, u_0)$$.

We are done.

• Nice! Can you outline how to prove Fact1? Commented Jun 23, 2022 at 14:37
• @SungjinKim Thanks. It is not nice. I will edit. Commented Jun 23, 2022 at 14:48

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 $$t\in[0.95,1]$$:

$$2\int_{0}^{1.252+5\left(t-1\right)}\frac{-\left(2-\left(2(1-c)+a^{b}c(2-c)-ac(1-c)\ln a\right)-x^{4}t-1\right)}{\cosh\left(\pi x\right)}dx+2\int_{1.252+5\left(t-1\right)}^{\infty}\frac{1}{\cosh\left(\pi x\right)^{2}}dx>\frac{e^{\frac{t}{2}}}{t+1}$$

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.

• I think the estimate $\cosh x \le e^{x^2 \over 2}$ is not enough (too loose). You can check $t = 1/2$. Commented Jun 20, 2022 at 15:02

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, $$e^{-\frac{t}{4}}$$.

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.

• Just mentioning that you can write \operatorname{erfc} to make it look like an operator. Commented Jun 18, 2022 at 16:10
• @MartinR Good point! Commented Jun 20, 2022 at 15:09