# Integral $\int_0^\infty e^{-x/2}x\log(1+kx^2)\,dx$

How to evaluate: $$\int_0^\infty e^{-x/2}x\log(1+kx^2)\,dx$$ Basically am evaluating value of $\log(1+c\chi^2)$ where $\chi^2$ is $\chi$-squared distributed

• Since $e^xx\log(1+kx^2)$ does not converge to zero, but actually diverges to infinity, this integral is fairly easy. – Thomas Andrews Jan 28 '15 at 4:00
• I am getting Expi() functions in between and am stuck of that – Jay Jan 28 '15 at 4:09
• So, it isn't obviously $-\infty$? Or did you want $\int_0^{X}$ and not $\int_{0}^\infty$? – Thomas Andrews Jan 28 '15 at 4:10
• If the question is wrong, fix the question, don't just add to the comments. – Thomas Andrews Jan 28 '15 at 4:36

Even if you would have meant to write $e^{-x}$, or $\displaystyle\int_{-\infty}^0$ , the integral would still not be expressible in terms of elementary functions and constants, since a simple substitution of the form $kx^2=\sinh^2t$ would immediately create an expression in terms of Bessel and Struve functions, and their various derivatives.
• Notice that the integral evaluates to $I'\bigg(-\dfrac12\bigg)$, where $I(\alpha)$ is the expression defined in this picture, with $k=\beta^2>0$. – Lucian Jan 28 '15 at 5:00