Change the order of integral.
$ \displaystyle \int_0^1\int_x^1 f(t) \ dt \ dx = \int_0^1\int_0^t f(t) \ dx \ dt$
Recall Binet's second $\ln \Gamma$ formula:
$$\int_{0}^{\infty} \frac{\arctan (t z)}{e^{2\pi t}-1} \, dt = \frac{1}{2} \ln \Gamma \left(\frac{1}{z}\right) + \frac{1}{2}\left(\frac{1}{z}-\frac{1}{2}\right) \ln (z) + \frac{1}{2z} - \frac{1}{4} \ln (2 \pi)$$
Consider the following integral:
$$\int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz = \frac{t z}{2 (1+t^2 z^2)(e^{2 \pi t}-1)} + \frac{\arctan (t z)}{2 (e^{2 \pi t}-1)}$$
Since $$\frac{t z}{2 (1+t^2 z^2)(e^{2 \pi t}-1)} = \frac{z}{2} \frac{\partial}{\partial z} \left(\frac{\arctan (t z)}{e^{2 \pi t}-1}\right)$$
$$\implies \int_{0}^{\infty} \int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz \, dt = \int_{0}^{\infty} \frac{z}{2} \frac{\partial}{\partial z} \left(\frac{\arctan (t z)}{e^{2 \pi t}-1}\right) \, dt+ \int_{0}^{\infty} \frac{\arctan (t z)}{2 (e^{2 \pi t}-1)} \, dt$$
Using Binet's formula we determine then:
$$\int_{0}^{\infty} \int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz \, dt = \frac{1}{4z}-\frac{1}{8}\ln(2\pi z)-\frac{\psi\left(\frac{1}{z}\right)}{4z}+\frac{1}{4}\ln\left(\Gamma\left(\frac{1}{z}\right)\right)-\frac{1}{8}$$
where $\psi$ is the digamma function.
Taking the derivative with respect to $z$ then the limit as $z \to 1$ we determine:
$$\boxed{\int_{0}^{\infty} \frac{x}{(e^{2 \pi x}-1)(x^2+1)^2} \, dx = \frac{\pi^2}{24} - \frac{3}{8}}$$
Best Answer
This integral is a classic example where the order of integration is important.
Your integral can be easily evaluated by rewriting it as
$$\int_0^{\infty}\frac{e^{-y/2}}{y}\Bigg[\int_0^y dx \Bigg]dy$$