Two curious "identities" on $x^x$, $e$, and $\pi$

You made a very nice observation! Often it is important to make a good guess than just to solve a prescribed problem. So it is surprising that you made a correct guess, especially considering the complexity of the formula.

I found a solution to the second integral in here, and you can also find a solution to the first integral at the link of this site.

Supplementary calculation of residue of the function

$$f(z)=e^{z\frac{e^z}{e^z-1}}\frac{e^z}{(e^z-1)^3}$$

at its triple pole $z=0$:

$f(z)$ is an odd function.Then

$$g(z)=f(z)^{\frac{1}{3}}=e^{\frac{1}{3}z(\frac{e^z}{e^z-1}+1)}\frac{1}{e^z-1}$$

is an odd function as well.

Hence $$g(z)=\frac{A_0}{z}+A_1 z+\cdots$$

and

$$Resf(z)_{\vert z=0}=3A_0^2A_1.$$

$z=0$ is a simple pole,then we can get $A_0=e^{\frac{1}{3}}$ without hesitation.

$$\frac{1}{e^z-1}=\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+\cdots=\frac{a_0}{z}+a_1+a_2 z+\cdots$$

$$\frac{z}{1-e^{-z}}+z=1+\frac{3}{2}z+\frac{1}{12}z^2+\cdots=b_0+b_1z+b_2 z^2+\cdots$$

Hence

$$\exp{(b_0+b_1z+b_2 z^2+\cdots)}=\exp(b_0)+b_1\exp(b_0)z+(b_1^2/2+b_2)\exp(b_0)z^2+\cdots,$$

i.e.,

$$e^{\frac{1}{3}z(\frac{e^z}{e^z-1}+1)}=e^{\frac{1}{3}}+\frac{1}{2}e^{\frac{1}{3}}z+\frac{11}{72}e^{\frac{1}{3}}z^2+\cdots$$

And $$A_1=\frac{1}{12}e^{\frac{1}{3}}-\frac{1}{2}\frac{1}{2}e^{\frac{1}{3}}+\frac{11}{72}e^{\frac{1}{3}}=-\frac{1}{72}e^{\frac{1}{3}}$$

Therefore $Resf(z)_{\vert z=0}=3A_0^2A_1=-e/24$.