Using the Abel–Plana summation formula with $f(x)=(z+x)^{-s}$ (principal branches), evaluate the integral $I=\int_0^\infty \frac{\arctan t}{e^{2\pi t}-1}\,dt$ in closed form.

What Abel–Plana is doing here

Applying Abel–Plana to $f(x)=(z+x)^{-s}$ produces an integral representation of the Hurwitz zeta function $$\zeta(s, z)=\sum_{n=0}^\infty (n+z)^{-s}, \qquad \Re(z)>0,$$ where the “correction term” is an integral involving the difference $f(it)-f(-it)$. When you rewrite that difference using principal arguments, an $\arctan$ appears.

From Abel–Plana to the Hurwitz zeta integral term

Abel–Plana gives (for $\Re(s)>1$, then by analytic continuation) $$\zeta(s, z)=\frac{z^{-s}}2+\int_0^\infty (z+x)^{-s}\, dx+i\int_0^\infty \frac{(z+it)^{-s}-(z-it)^{-s}}{e^{2\pi t}-1}\, dt.$$ Also, $$\int_0^\infty (z+x)^{-s}\, dx=\frac{z^{1-s}}{s-1}.$$

Differentiate at $s=0$ to get $\log\Gamma$ (Binet’s formula)

A standard identity linking Hurwitz zeta and the Gamma function is $$\zeta'(0, z)=\log\Gamma(z)-\frac12\log(2\pi).$$ Differentiate the Abel–Plana representation at $s=0$. The key point for the correction integral is the principal-branch evaluation $$\log(z\pm it)=\tfrac12\log(z^2+t^2)\pm i\arctan\!\Big(\frac{t}{z}\Big), \qquad \Re(z)>0,$$ so the imaginary parts combine to give an $\arctan$ term. This yields Binet’s second formula: $$\log\Gamma(z)=\Big(z-\tfrac12\Big)\log z-z+\tfrac12\log(2\pi)+2\int_0^\infty \frac{\arctan(t/z)}{e^{2\pi t}-1}\, dt, \qquad \Re(z)>0.$$

Plug in $z=1$ to match the given integral

Set $z=1$. Then $\Gamma(1)=1$, so $\log\Gamma(1)=0$, and $\log 1=0$. Binet’s formula becomes $$0=-1+\tfrac12\log(2\pi)+2\int_0^\infty \frac{\arctan t}{e^{2\pi t}-1}\, dt.$$ Therefore $$2I=1-\tfrac12\log(2\pi)\quad\Rightarrow\quad I=\frac12-\frac14\log(2\pi).$$