WolframAlpha yields
$$\int_1^\infty u^{-\pi} (u+1)^{-1} du = \frac{1}{2}\psi_0(0.5 + 0.5 π) - \frac{1}{2}\psi_0(0.5π)$$
but does not give an explanation. Does anyone know how to prove this?
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Do you know what function $\psi_0$ stands for ? – Jean Marie Feb 22 '20 at 11:48
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$\psi_0(x)=\psi(x)$ is the polygamma function. – Jenny Reininger Feb 22 '20 at 11:50
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2I have taken the liberty to change your previous title "Wolfram Alpha gives" which was uninformative – Jean Marie Feb 22 '20 at 11:50
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It is likely that the definition $\psi(x)=\Gamma'(x)/\Gamma(x)$ has to be used... – Jean Marie Feb 22 '20 at 12:03
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$$\lim_{b\to 0}\big(\Gamma(b)-\mathrm{B}(a,b)\big)=\lim_{b\to 0}\frac{\Gamma(b+1)}{\Gamma(a+b)}\frac{\Gamma(a+b)-\Gamma(a)}{b}=\frac{\Gamma'(a)}{\Gamma(a)}=\psi(a)$$ assuming $a,b>0$, which implies $$\psi(b)-\psi(a)=\lim_{c\to 0}\big(\mathrm{B}(a,c)-\mathrm{B}(b,c)\big)=\int_0^1\frac{x^{a-1}-x^{b-1}}{1-x}~dx.$$ The given integral transforms into a particular case of the above after substituting $u=1/\sqrt{x}$.
metamorphy
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(... and noting that $\frac{1}{1+\sqrt{x}}=\frac{1-\sqrt{x}}{1-x}$.) – metamorphy Feb 22 '20 at 12:20