Question Number 214051 by issac last updated on 25/Nov/24 | ||
$$\mathrm{evaluate}\:\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}...\: \\ $$$$\mathrm{i}\:\mathrm{use}\:\mathrm{Feynman}'\mathrm{s}\:\mathrm{trick}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{integral} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\pi} \:{e}^{{t}\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$${I}^{\left(\mathrm{1}\right)} \left({t}\right)=\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{sin}^{\mathrm{2}} \left({u}\right){e}^{{t}\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left({u}\right)\right){e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}−\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{cos}^{\mathrm{2}} \left({u}\right){e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}−\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{sin}^{\mathrm{2}} \left({u}+\frac{\pi}{\mathrm{2}}\right){e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$${u}+\frac{\pi}{\mathrm{2}}\:={w}\:\:\rightarrow\:\:\mathrm{d}{u}=\mathrm{d}{w} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \left({w}\right){e}^{{t}\centerdot\mathrm{cos}^{\mathrm{2}} \left({w}\right)} \:\mathrm{d}{w}= \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\:\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \left({w}\right){e}^{{t}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left({w}\right)\right)} \mathrm{d}{w} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\:\mathrm{sin}^{\mathrm{2}} \left({w}\right){e}^{{t}} {e}^{−{t}\mathrm{sin}^{\mathrm{2}} \left({w}\right)} \mathrm{d}{w}={e}^{{t}} \int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \left({w}\right){e}^{−{t}\mathrm{sin}^{\mathrm{2}} \left({w}\right)} \mathrm{d}{w} \\ $$$${I}^{\left(\mathrm{1}\right)} \left({t}\right)=\int_{\mathrm{0}} ^{\:\pi} \:{e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}−{e}^{{t}} \int_{\mathrm{0}} ^{\:\pi} \:\mathrm{sin}^{\mathrm{2}} \left({u}\right){e}^{−{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:...\mathrm{anymore}...\mathrm{help} \\ $$ | ||
Commented by Ghisom last updated on 26/Nov/24 | ||
$$\mathrm{Feynman}\:\mathrm{can}'\mathrm{t}\:\mathrm{help}\:\mathrm{here} \\ $$ | ||
Answered by Ghisom last updated on 26/Nov/24 | ||
$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{e}^{\mathrm{sin}^{\mathrm{2}} \:{x}} {dx}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{e}^{\mathrm{sin}^{\mathrm{2}} \:{x}} {dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2}{x}+\pi\right] \\ $$$$=\sqrt{\mathrm{e}}\underset{\pi} {\overset{\mathrm{2}\pi} {\int}}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{t}} {dt}=\sqrt{\mathrm{e}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{t}} {dt} \\ $$$$\mathrm{Modified}\:\mathrm{Bessel}\:\mathrm{Function}\:\mathrm{of}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{Kind} \\ $$$${I}_{{n}} \left({z}\right)=\frac{\mathrm{1}}{\pi}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{e}^{{z}\mathrm{cos}\:\theta} \mathrm{cos}\:\left({n}\theta\right)\:{d}\theta \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:{n}=\mathrm{0}\wedge{z}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\sqrt{\mathrm{e}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{t}} {dt}=\sqrt{\mathrm{e}}\pi{I}_{\mathrm{0}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$ | ||
Commented by issac last updated on 27/Nov/24 | ||
$${wow}\:{thx} \\ $$ | ||