Question Number 218043 by SdC355 last updated on 27/Mar/25
$$\int\:\:{e}^{−{st}} {J}_{\nu} \left({t}\right){dt}=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}.\:\:\nu\in\mathbb{R}^{+} \\ $$$$\frac{\mathrm{d}\:\:}{\mathrm{d}{s}}\int\:\:{e}^{−{st}} {J}_{\nu} \left({t}\right){dt}=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} \left({s}+\nu\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)}{\:\sqrt{\left({s}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }} \\ $$$$\mathrm{and}\:\frac{\mathrm{d}\:\:}{\mathrm{d}{s}}\int\:\:{e}^{−{st}} {J}_{\nu} \left({t}\right){dt}=−\int\:\:{te}^{−{st}} {J}_{\nu} \left({t}\right){dt} \\ $$$$\int\:\:{te}^{−{t}} {J}_{−\frac{\mathrm{1}}{\mathrm{2}}} \left({t}\right){dt}=−\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}^{\mathrm{3}} }} \\ $$
Answered by SdC355 last updated on 27/Mar/25
$$\mathrm{Q218025} \\ $$