I am trying to find the expected value of a probability density function. Solving the integral of the function times its random variable with integration by parts, I arrive at the following integrals which are rather complex. I'd appreciate very much if you could provide some directions on how I could tackle this problem. Thanks!
$$\int_0^\infty \ln(2x+1) \, x e^{-2Tx} \, dx + \int_0^\infty \ln(2x+1) \, e^{-2Tx} \, dx$$
Sub $x=(u-1)/(2 N)$; then the integral is equal to
$$\frac1{2 N} e^{T/N} \int_1^{\infty} \frac{du}{u} e^{-T u/N} = \frac1{2 N} e^{T/N} \Gamma \left (0,\frac{T}{N} \right )$$
where $\Gamma$ is the upper incomplete gamma function.