Question Number 142362 by mnjuly1970 last updated on 30/May/21
$$ \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:\:{that}: \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} {ln}\left(\frac{\mathrm{1}}{{x}}\right).{j}_{\mathrm{0}} \left({x}\right){dx}:=\:\gamma+{ln}\left(\mathrm{2}\right)\: \\ $$$$\:\:\:\:\:{Hint}:\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:{j}_{\mathrm{0}} \left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\mathrm{2}^{\mathrm{2}{n}} .\Gamma^{\mathrm{2}} \left({n}+\mathrm{1}\right)}\:\left({Bessel}\:{function}\right) \\ $$$$\:\:\:\:{Hint}:\mathrm{2}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathscr{L}\:\left[\:{j}_{\mathrm{0}} \left({x}\right)\right]=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }} \\ $$
Answered by Kamel last updated on 30/May/21
Commented by mnjuly1970 last updated on 31/May/21
$${thanks}\:{alot}\:{mr}\:{kamel} \\ $$