Happy New Year! Using the Abel-Plana formula on the Basel sum, I found that $$-i\frac{\pi^2-9}6=\int_0^\infty\frac1{(ti+1)^2(e^{2\pi t}-1)}-\frac1{(-ti+1)^2(e^{2\pi t}-1)}dt$$ How do I solve this integral without the knowledge of the Abel Plana summation formula?
Calculus – Solving Complex Improper Integrals
calculusintegration
Best Answer
Taking the common denominator of the expression in RHS, we get:
$$I = \int_0^{\infty} \frac{1}{e^{2\pi t}-1} \cdot \frac{-4it}{(1+t^2)^2} dt = -i\cdot \int_0^{\infty} \frac{4tdt}{(1+t^2)^2\cdot (e^{2\pi t} -1)}$$
Now, from Binet's second formula for the digamma function, we have:
$$\psi(z)=\log(z)-\frac{1}{2z}-2\int_0^\infty \frac{x}{(x^2+z^2)(e^{2\pi x}-1)}dx$$
Differentiating the equation once with respect to $z$, we get
$$\psi^1(z) = \frac{1}{z} + \frac{1}{2z^2} + z\cdot \int_0^{\infty} \frac{4x}{(x^2+z^2)^2\cdot (e^{2\pi x} -1)} dx$$
For $z=1$, we get
$$\psi^1(1) = \frac{\pi^2}{6} = \frac{3}{2} + \int_0^{\infty} \frac{4xdx}{(1+x^2)^2\cdot (e^{2\pi x} -1)}$$
Thus we get
$$\int_0^{\infty} \frac{4xdx}{(1+x^2)^2\cdot (e^{2\pi x}-1)} = \frac{\pi^2-9}{6}$$
And thus we arrive at the result:
$$ I = -i\cdot \frac{\pi^2-9}{6}$$