Question Number 222922 by MrGaster last updated on 10/Jul/25
$$ \\ $$$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} {J}_{\mathrm{0}} \left(\mathrm{ln}\frac{\mathrm{1}}{{x}}\right){dx}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Answered by MrGaster last updated on 11/Jul/25
$${u}=\mathrm{ln}\frac{\mathrm{1}}{{x}}\Rightarrow{x}={e}^{−{u}} ,{dx}=−{e}^{−{u}} {du} \\ $$$${x}=\mathrm{0}\Rightarrow{u}=\infty,{x}=\mathrm{1}\Rightarrow{u}=\mathrm{0} \\ $$$$=\int_{\infty} ^{\mathrm{0}} {J}_{\mathrm{0}} \left({u}\right)\left(−{e}^{−{u}} \right){du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {J}_{\mathrm{0}} \left({u}\right){e}^{−{u}} {du} \\ $$$${J}_{\mathrm{0}} \left({u}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}!\right)^{\mathrm{2}} }\left(\frac{{u}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{u}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}!\right)^{\mathrm{2}} \mathrm{4}^{{k}} }{u}^{\mathrm{2}{k}} {du} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}!\right)^{\mathrm{2}} \mathrm{4}^{{k}} }\int_{\mathrm{0}} ^{\infty} {u}^{\mathrm{2}{k}} {e}^{−{u}} {du} \\ $$$$\int_{\mathrm{0}} ^{\infty} {u}^{\mathrm{2}{k}} {e}^{−{u}} {du}=\Gamma\left(\mathrm{2}{k}+\mathrm{1}\right)=\left(\mathrm{2}{k}\right)! \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \left(\mathrm{2}{k}\right)!}{\left({k}!\right)^{\mathrm{2}} \mathrm{4}^{{k}} } \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{k}}\\{{k}}\end{pmatrix}\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} \\ $$$$=\left(\mathrm{1}−\mathrm{4}\centerdot\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\left(\mathrm{1}+\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$