Calculation of some integrals 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?
 A: 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 $x=0$!
