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The next functions are defined: $$ f(y)=\frac{1}{1+e^{-2y}} \\ g_1(z)=\frac{1}{1+z^2},\quad g_2(z)=e^{-z^2},\quad g_3(z)=\frac{1}{cosh(z)},\quad g_4(z)=\frac{sin(z)}{z}$$ Is there a way to calculate the integral $$\int_{0}^{\infty}f(y)g_i(x-y)dy $$ for all values of $x \in \mathbb{R}$ for some $i=1,\dots,4$. So, I need to find just one function $g_i(x)$ such that the integral defined above can be calculated explicitly. I have tried some simple methods of integrating and the Laplace transformation, but none of them helped much. I suspect that this question can not be answered positively.

Edited:

There is an answer for $g_3(z)$. I thought that another result for the function $$g_{3,modif}(z)=\frac{1}{cosh(kz)} $$ would easily follow from the origin result. I was wrong. Mathematica seems to calculate the integral with $g_{3,modif}(z)$ for various $k>0$, but the higher the value of k the more time it takes to calculate. In case the answer in explicit form couldn't be obtained is there a subsequence ${k_j}$ such that the appropriate answer exists?

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    $\begingroup$ Is $g_4(z)$ meant to be $\frac{\sin(z)}{z}$? $\endgroup$ – Robert Howard Dec 3 '18 at 16:07
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    $\begingroup$ Entering $\int_{0}^{\infty}f(y)g_3(x-y)dy$ on integral-calculator.com gives an answer. $\endgroup$ – greelious Dec 3 '18 at 16:25
  • $\begingroup$ Indeed. I hoped that having the result for $g_3(z)$ I could get easily get another result with $g_{3,modif}=\frac{1}{cosh(kz)}$ for k>0, but I was wrong. Thought, Mathematica calculates distinct integrals for even values of k. The higher the value of k the more time calculation lasts. $\endgroup$ – Artem Zefirov Dec 3 '18 at 17:01
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The third can be calculated explicitly. One might be able to guess this because it's expression is most closely related to $f$ since $$g_3(z) = \frac{1}{\cosh(z)} = \frac{2}{e^z + e^{-z}} = \frac{2e^{-z}}{1+e^{-2z}} = 2e^{-z}f(z)$$ so we can write $$\int_0^\infty f(z)g_3(x-z)dz = \int_0^\infty \frac{e^{2z}}{1+e^{2z}}\frac{2e^{z}e^x}{e^{2x} + e^{2z}}dz $$ Now we can make the substitution $u = e^z$ to get $$2e^x\int_{-\infty}^\infty \frac{u^2}{(1+u^2)(e^{2x}+u^2)}du$$ And now (if $x \neq 0$) we can do a partial fraction decomposition to get $$2e^x\frac{1}{1-e^{2x}}\left( \int_{-\infty}^\infty \frac{1}{1+u^2} - \frac{e^{2x}}{e^{2x}+u^2}du \right)$$ And this is easily evaluated with arctangents as $$2e^x\frac{1}{1-e^{2x}}\left(\pi - e^x \pi\right) = 2\pi\frac{e^x}{1+e^x}$$ And finally if $x=0$, then we must find $$\int_{-\infty}^\infty \frac{u^2}{(1+u^2)^2}du $$ And here we can use the substitution $u=\tan(\theta)$ to get $$\int_{-\pi/2}^{\pi/2} \frac{\tan^2(\theta)}{\sec^4(\theta)}\sec^2(\theta)d\theta = \int_{-\pi/2}^{\pi/2} \sin^2(\theta)d\theta = \pi$$ So the expression we derived for general $x$ does in fact also turn out to work for $x=0$!

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  • $\begingroup$ Thanks. Your idea is clear in spite of some mistakes in the integral limits and the numerator value after the substitution. $\endgroup$ – Artem Zefirov Dec 4 '18 at 12:38
  • $\begingroup$ I wiil attempt an edit. Thanks for pointing out mistakes! $\endgroup$ – Isaac Browne Dec 4 '18 at 14:53

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