Show that $\int_0^{\infty} (\exp(x) - 2^{\sqrt 5 +1} + 1) (\coth(x) - 1) x^{\sqrt 5} dx = 0 $

Question

Show that

$$ \int_0^{\infty} (\exp(x) - 2^{\sqrt 5 +1} + 1) (\coth(x) - 1) x^{\sqrt 5} dx = 0 $$

I wonder how many distinct methods there are.

Answer 1

Using identity:

$$\sum _{n=1}^{\infty } \exp (-n 2 x)=\frac{1}{-1+e^{2 x}}$$ $$\int_0^{\infty } t^{x-1} \exp (-a t) \, dt=a^{-x} \Gamma (x)$$ $$\sum _{n=1}^{\infty } n^{-x}=\zeta (x)$$ so:

$\int_0^{\infty } \left(\exp (x)-2^{\sqrt{5}+1}+1\right) (\coth (x)-1) x^{\sqrt{5}} \, dx=\\\int_0^{\infty } \frac{\left(-1+2^{1+\sqrt{5}}-e^x\right) x^{\sqrt{5}}}{-1+e^{2 x}} \, dx=\\\sum _{n=1}^{\infty } \int_0^{\infty } \left(-1+2^{1+\sqrt{5}}-e^x\right) x^{\sqrt{5}} \exp (-n 2 x) \, dx=\sum _{n=1}^{\infty } \left(\int_0^{\infty } -x^{\sqrt{5}} \exp (-n 2 x) \, dx+\int_0^{\infty } 2^{1+\sqrt{5}} x^{\sqrt{5}} \exp (-n 2 x) \, dx-\int_0^{\infty } \exp (x) x^{\sqrt{5}} \exp (-n 2 x) \, dx\right)=\\\sum _{n=1}^{\infty } \left(n^{-1-\sqrt{5}} \Gamma \left(1+\sqrt{5}\right)-2^{-1-\sqrt{5}} n^{-1-\sqrt{5}} \Gamma \left(1+\sqrt{5}\right)-(-1+2 n)^{-1-\sqrt{5}} \Gamma \left(1+\sqrt{5}\right)\right)=\sum _{n=1}^{\infty } n^{-1-\sqrt{5}} \Gamma \left(1+\sqrt{5}\right)-\sum _{n=1}^{\infty } 2^{-1-\sqrt{5}} n^{-1-\sqrt{5}} \Gamma \left(1+\sqrt{5}\right)-\sum _{n=1}^{\infty } (-1+2 n)^{-1-\sqrt{5}} \Gamma \left(1+\sqrt{5}\right)=\\\Gamma \left(1+\sqrt{5}\right) \zeta \left(1+\sqrt{5}\right)-2^{-1-\sqrt{5}} \Gamma \left(1+\sqrt{5}\right) \zeta \left(1+\sqrt{5}\right)-\Gamma \left(1+\sqrt{5}\right) \zeta \left(1+\sqrt{5}\right)+2^{-1-\sqrt{5}} \Gamma \left(1+\sqrt{5}\right) \zeta \left(1+\sqrt{5}\right)=0$