# Inequality between Integral expressions

I want to prove that there exists a costant $C_{t,l}$ depending only on $t\in \mathbb{R}^+$ and $l\in \mathbb{Z}^+$ such that, for any $d\in \mathbb{R}^+$, the following inequality of "Heat Kernels" holds:

$$\int_d^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-\cosh(d)}} e^{lx}dx\leq C_{t,l}\int_d^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-\cosh(d)}} dx$$

Please note that the only difference between the integrals is the factor $e^{lx}$ in the LHS.

My main attempt consisted in defining $C_{t,l}$ in the following way:

$$C_{t,l}:=\frac{\int_0^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-1}} e^{lx}dx}{\int_0^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-1}} dx},$$

i.e. by forcing the above inequality to be an equality for the limit case $d=0$. The well definition of $C_{t,l}$ follows by the fact that for $d$ approaching $0$ the additional factor $e^{lx}$ is integrated over regions where it is "hardly" bounded.

So I'm to prove that:

$$\frac{\int_d^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-\cosh(d)}} e^{lx}dx}{\int_d^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-\cosh(d)}} dx}\leq\frac{\int_0^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-1}} e^{lx}dx}{\int_0^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-1}} dx}.$$

Heuristically this is clear. Assuming $d<1$, on the LHS I'm not considering regions where the "weight" of the additional factor $e^{lx}$ is maximum. Unfortunately I don't know how to prove it.

Do you have any suggestions? Please note that I'm interested only in the existence of $C_{t,l}$ and not in finding an expression for it. In specific feel free to use (and share!) all the shortcuts you know!

Thank you very much!

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Such a constant $C_{t,l}$ does not exist since $$\int_d^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-\cosh(d)}} e^{lx}dx\geq e^{dl}\int_d^\infty \frac{xe^{-\frac{x^2}{4t}}}{\sqrt{\cosh(x)-\cosh(d)}} dx.$$