Jtdylktuy

Motivated by this paper.

Conjecture:

$$int_{-infty}^{+infty}{ke^xpm1over pi^2+(e^x-x+1)^2}cdot{(e^x+1)^2over pi^2+(e^x+x+1)^2}cdot 2x ,mathrm dx=k,tag1$$
where $k$ is a real number.

Making an attempt:

$u=e^x+1implies ,mathrm du=e^x,mathrm dx$ and let $k=1$ for simplification, then (1) becomes

$$int_{1}^{infty}{u^3over pi^2+(u-x)^2}cdot{ln(u-1)over pi^2+(u+x)^2}cdot{2mathrm duover u-1}.tag2$$

I have no idea where to go from here! I don't think substitution work here, probably using contour integration.

How can we prove (1)?

Some integrals

• Let us prove that

$$boxed{I_0 = intlimits_{-infty}^{+infty}{dzoverleft(e^z-z+1right)^2+pi^2} = {1over2}}$$
Roots of the denominator can be defined from the system
$$begin{cases}
z=x+iy\
left(e^xcos y - x + 1 + ie^xsin y - iyright)^2 + pi^2 = 0,
end{cases}$$
$$begin{cases}
z=x+iy\
left(e^xcos y - x + 1right)^2 - left(e^xsin y - yright)^2 + pi^2 = 0\
left(e^xcos y - x + 1right)left(e^xsin y - yright) = 0,
end{cases}$$
$$begin{cases}
z=x+iy\
e^xcos y = x - 1\
left|e^xsin y - yright| = pi,
end{cases}$$
with the solutions $z=pmpi i$ (see also Wolfram Alpha).

So,
$$I_0 = 2pi i,mathrm{Res}_{z=pi i}{1overleft(e^z-z+1right)^2+pi^2} = 2pi ilim_{ztopi i}{1over2left(e^z-z+1right)left(e^z-1right)} = {1over2}.$$

• Let us prove that

$$boxed{I_1 = intlimits_{-infty}^{+infty}{dzoverleft(e^z+z+1right)^2+pi^2} = {2over3}}$$
Roots of the denominator can be defined from the system
$$begin{cases}
z=x+iy\
left(e^xcos y + x + 1 + ie^xsin y + iyright)^2 + pi^2 = 0,
end{cases}$$
$$begin{cases}
z=x+iy\
left(e^xcos y + x + 1right)^2 - left(e^xsin y + yright)^2 + pi^2 = 0\
left(e^xcos y + x + 1right)left(e^xsin y + yright) = 0,
end{cases}$$
$$begin{cases}
z=x+iy\
e^xcos y + x + 1 = 0\
left|e^xsin y + yright| = pi,
end{cases}$$
with the solutions $z=pmpi i$ (see also Wolfram Alpha).

Note that the point $z=pi i$ is a second-order pole, so
$$I_1 = 2pi i,mathrm{Res}_{z=pi i}{1overleft(e^z+z+1right)^2+pi^2} = 2pi ilim_{ztopi i} {dover dz}left({(z-pi i)^2overleft(e^z+z+1right)^2+pi^2}right) = {2over3}.$$
(see also Wolfram Alpha).

• Let us prove that

$$boxed{I_2 = intlimits_{-infty}^{+infty}{e^zdzoverleft(e^z-z+1right)^2+pi^2} = {1over2}}$$

Really,
$$I_2 = intlimits_{-infty}^{+infty}{e^zdzoverleft(e^z-z+1right)^2+pi^2}= intlimits_{-infty}^{+infty}{e^z-1overleft(e^z-z+1right)^2+pi^2},dz + I_0$$
$$ = {1overpi}left.arctan{e^z-z-1overpi}right|_{-infty}^{+infty} + {1over 2} = {1over2}.$$

• Let us prove that

$$boxed{I_3 = intlimits_{-infty}^{+infty}{e^zdzoverleft(e^z+z+1right)^2+pi^2} = {1over3}}$$

Similarly,
$$I_3 = intlimits_{-infty}^{+infty}{e^zdzoverleft(e^z+z+1right)^2+pi^2}= intlimits_{-infty}^{+infty}{e^z+1overleft(e^z+z+1right)^2+pi^2},dz - I_1$$
$$ = {1overpi}left.arctan{e^z+z-1overpi}right|_{-infty}^{+infty} - {2over 3} = {1over3}.$$

• Let us prove that

$$boxed{I_4 = intlimits_{-infty}^{+infty}{2z(e^z+1)^2overleft(left(e^z-z+1right)^2+pi^2right)left(left(e^z+z+1right)^2+pi^2right)},dx = 0}$$

Really,
$$I_4 = intlimits_{-infty}^{+infty}{2z(e^z+1)^2overleft(left(e^z-z+1right)^2+pi^2right)left(left(e^z+z+1right)^2+pi^2right)},dx$$
$$= intlimits_{-infty}^{+infty}{e^z+1over2}left({1overleft(e^z-z+1right)^2+pi^2} - {1overleft(e^z+z+1right)^2+pi^2}right),dx$$
$$= {I_2+I_0-I_3-I_1over2} = {1over2}left({1over2}+{1over2}-{2over3}-{1over3}right) = 0.$$

• Let us prove that

$$boxed{I_5 = intlimits_{-infty}^{+infty}{2ze^z(e^z+1)^2overleft(left(e^z-z+1right)^2+pi^2right)left(left(e^z+z+1right)^2+pi^2right)},dx =
1}$$

The denominator is
$$D(z) = left(left(e^z+1right)^2+z^2
+pi^2 - 2zleft(e^z+1right)right) left(left(e^z+1right)^2+z^2+pi^2 + 2zleft(e^z+1right)right)$$
$$= left(left(e^z+1right)^2+z^2+pi^2right)^2 - 4z^2left(e^z+1right)^2,$$
$$D'(z) = 4left(e^z+z+1right)left(left(e^z+1right)^2+z^2+pi^2right) -8zleft(e^z+1right)left(e^z+z+1right)$$
$$=4left(e^z+z+1right)left(left(e^z-z+1right)^2+pi^2right)$$

The point $z=pi i $ is the simple pole. So,

$$I_5 = 2pi i,mathrm{Res}_{z=pi i}{2ze^z(e^z+1)^2overleft(left(e^z-z+1right)^2+pi^2right)left(left(e^z+z+1right)^2+pi^2right)}$$
$$ = 2pi i,lim_{ztopi i}{2ze^z(e^z+1)^2over D'(z)} = 1.$$
(see also Wolfram Alpha)

Final calculations

$$I = intlimits_{-infty}^{+infty}{ke^xpm1over pi^2+(e^x-x+1)}cdot{(e^x+1)^2over pi^2+(e^x+x+1)^2}cdot 2x mathrm dx$$
$$= kI_5pm I_4 = k.$$

Finally,
$$boxed{boxed{I = k}}$$

First note that considering

$$F(k)=int_{-infty}^{+infty}{(ke^xpm1)over pi^2+(e^x-x+1)^2}cdot{(e^x+1)^2over pi^2+(e^x+x+1)^2}cdot 2x mathrm dx$$

Let $x to log(x)$

$$F(k)=int_{0}^{+infty}{(kxpm1) over pi^2+(x-log(x)+1)^2}cdot{(x+1)^2over pi^2+(x+log(x)+1)^2}cdot frac{2log(x)}{x} mathrm dx = k$$

By separating the integrals note that

$$I_1=int_{0}^{+infty}{1 over pi^2+(x-log(x)+1)^2}cdot{(x+1)^2over pi^2+(x+log(x)+1)^2}cdot frac{2log(x)}{x} mathrm dx=0$$

I could prove it numerically using Matlab. Hence I only show

$$I_2=int_{0}^{+infty}{log(x) over pi^2+(x-log(x)+1)^2}cdot{(x+1)^2over pi^2+(x+log(x)+1)^2}cdot mathrm dx = frac{1}{2}$$

Consider the function

$$f(z) = frac{(z-1)^2}{(1-(z+log z))(1-(z-log(z))}$$

Integrated around a key-hole contour around the principle branch of the logarithm

$$log(z) = log|z|+imathrm{Arg}(z)$$

Hence the contour

By taking the limits the smaller circle and the bigger one go to zero hence

$$int_{-infty}^{0}frac{(x-1)^2}{(1-(x+log|x|+ipi ))(1-(x-log|x|-ipi)}dx+int_{0}^{-infty}frac{(x-1)^2}{(1-(x+log|x|-ipi ))(1-(x-log|x|+ipi)}dx = 2pi imathrm{Res}(f,1)$$

Convert to the positive limit

$$int_{0}^{infty}frac{(x+1)^2}{(1+x-log x-ipi )(1+x+log x+ipi)}-frac{(x+1)^2}{(1+x-log x+ipi )(1+x+log x-ipi)}dx = 2pi imathrm{Res}(f,1)$$

This magically reduces to our integral

$$int_{0}^{+infty}{4pi ,i log(x) over pi^2+(x-log(x)+1)^2}cdot{(x+1)^2over pi^2+(x+log(x)+1)^2}cdot mathrm dx = 2pi imathrm{Res}(f,1)$$

Note that

$$mathrm{Res}(f,1) = lim_{z to 1}frac{(z-1)^3}{(1-(z+log z))(1-(z-log(z))} = 1$$

Hence we finally get our result

$$int_{0}^{+infty}{log(x) over pi^2+(x-log(x)+1)^2}cdot{(x+1)^2over pi^2+(x+log(x)+1)^2}cdot mathrm dx = frac{1}{2}$$

Using the same approach we could show

$$int^infty_{-infty}frac{dx}{(e^x-x+1)^2+pi^2}=frac{1}{2}$$