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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

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

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    $\begingroup$ By using numerical methods based on their fast-converging series expansion. $\endgroup$ – Lucian Jan 28 '15 at 4:32
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    $\begingroup$ Notice that the integral evaluates to $I'\bigg(-\dfrac12\bigg)$, where $I(\alpha)$ is the expression defined in this picture, with $k=\beta^2>0$. $\endgroup$ – Lucian Jan 28 '15 at 5:00

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