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

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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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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.