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 $t\in[0,1]$?

  • $\begingroup$ 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? $\endgroup$
    – fedja
    Commented Jun 10, 2022 at 4:14
  • $\begingroup$ @fedja If your whole proof simple? (I also have a proof assuming that it is not difficult to verify something like $\int_0^\infty \frac{\mathrm{e}^{-x^2-1/2}}{\cosh \pi x}\,\mathrm{d} x > \frac14, \int_0^\infty \frac{x^2\mathrm{e}^{-x^2-1/2}}{\cosh \pi x}\,\mathrm{d} x > \frac{1}{30}$. $\endgroup$
    – River Li
    Commented Jun 10, 2022 at 6:30
  • $\begingroup$ @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. $\endgroup$
    – fedja
    Commented Jun 10, 2022 at 9:17
  • $\begingroup$ @fedja Perhaps no easy way. So your proof is nice (remains to check $t=1$). $\endgroup$
    – River Li
    Commented Jun 10, 2022 at 13:14
  • $\begingroup$ 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. $\endgroup$ Commented Jun 14, 2022 at 1:42

8 Answers 8


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

  • $\begingroup$ Following your solution, I see that $J'(t)\geq -1/4 + t/4$. So, we might have $J(t)\geq -t/4 + t^2/8$. $\endgroup$ Commented Jun 11, 2022 at 19:50
  • $\begingroup$ 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 $J''(0)$? $\endgroup$ Commented Jun 11, 2022 at 20:53
  • $\begingroup$ Upon closer inspection, it appears that $J'(t)$ is actually concave so the direction of the inequality is the other way round, unfortunately. $\endgroup$
    – TheSimpliFire
    Commented Jun 11, 2022 at 21:52
  • $\begingroup$ 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 $t\ge0$. $\endgroup$
    – TheSimpliFire
    Commented Jun 11, 2022 at 21:58
  • 2
    $\begingroup$ 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. $\endgroup$ 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 $\exp\left(-\frac t4 + \frac{t^2}{16}\right)$.

  • $\begingroup$ 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. $\endgroup$
    – TheSimpliFire
    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 $$2\int_0^\infty\frac{\mathrm{e}^{-tx^2}}{\cosh\pi x}\,\mathrm{d}x\ge\exp\left(-\frac t4+\frac{t^2}{16}\right).$$


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.

  • $\begingroup$ Nice! Can you outline how to prove Fact1? $\endgroup$ Commented Jun 23, 2022 at 14:37
  • $\begingroup$ @SungjinKim Thanks. It is not nice. I will edit. $\endgroup$
    – River Li
    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.

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


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.


$$f(t)= \sum_{k=0}^{\infty}(-1)^k e^{t(k+\frac{1}{2})^2}\operatorname{erfc\left [\sqrt{t}(k+\frac{1}{2}) \right ]}$$

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.

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

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