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Question Number 70237 by mathmax by abdo last updated on 02/Oct/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{xsin\left(\alpha x\right)}{1+x^4}dxwith\alpha real$$

Commented by mind is power last updated on 02/Oct/19

$$niceideleatmypostmistack[$$$$\frac{zsin\left(z\right)}{1+z^4}when\mid z\mid \to \mathrm{\infty }isnotdefind$$$$z={re}^{ia}a\in [0,\pi [$$$$sin\left(z\right)=\frac{e^{{ire}^{ia}}-e^{{ire}^{ia}}}{2i}$$$$thefactor{e}^{-{ire}^{ia}}=e^{-irvos\left(a\right)}.e^{rsin\left(a\right)}\mid e^{-{ire}^{ia}}\mid =e^{rsin\left(a\right)}witchisgetbigaswewant$$$$\Rightarrow limzf\left(z\right)\nRightarrow 0jordanlemmadontnotapplie$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz\ne 2i\pi resid(.....)$$$$ihavedoneitquicklywithoutseeingwhatgoingonSorrysir$$

Commented by mathmax by abdo last updated on 02/Oct/19

$$letA={\int }_0^{\mathrm{\infty }}\frac{xsin\left(\alpha x\right)}{x^4+1}dx\Rightarrow 2A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{xsin\left(\alpha x\right)}{x^4+1}dx$$$$=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x{e}^{i\alpha x}}{x^4+1}dx\right)let\phi \left(z\right)=\frac{z{e}^{i\alpha z}}{z^4+1}polesof\phi ?$$$$\phi \left(z\right)=\frac{z{e}^{i\alpha z}}{(z^2-i)(z^2+i)}=\frac{z{e}^{i\alpha z}}{(z-\sqrt{i})(z+\sqrt{i})(z-\sqrt{-i})(z+\sqrt{-i})}$$$$=\frac{z{e}^{i\alpha z}}{(z-e^{\frac{i\pi }{4}})(z+e^{\frac{i\pi }{4}})(z-e^{-\frac{i\pi }{4}})(z+e^{-\frac{i\pi }{4}})}sothepolesof\phi are$$$$\stackrel{-}{+}{e}^{\frac{i\pi }{4}}and\stackrel{-}{+}{e}^{-\frac{i\pi }{4}}and{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{Res(\phi ,e^{\frac{i\pi }{4}})+Res(\phi ,-e^{-\frac{i\pi }{4}})\}$$$$Res(\phi ,e^{\frac{i\pi }{4}})={lim}_{z\to e^{\frac{i\pi }{4}}}(z-e^{\frac{i\pi }{4}})\phi \left(z\right)$$$$=\frac{e^{\frac{i\pi }{4}}{e}^{i\alpha e^{\frac{i\pi }{4}}}}{2e^{\frac{i\pi }{4}}\left(2i\right)}=\frac{e^{i\alpha (\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})}}{4i}=\frac{1}{4i}{e}^{\frac{i\alpha }{\sqrt{2}}}{e}^{-\frac{\alpha }{\sqrt{2}}}$$$$Res(\phi ,-e^{-\frac{i\pi }{4}})={lim}_{z\to -e^{-\frac{i\pi }{4}}}(z+e^{-\frac{i\pi }{4}})\phi \left(z\right)$$$$=\frac{-e^{-\frac{i\pi }{4}}{e}^{i\alpha (-e^{-\frac{i\pi }{4}})}}{(-2i)(-2e^{-\frac{i\pi }{4}})}=\frac{1}{4i}{e}^{-i\alpha (\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})}=-\frac{1}{4i}{e}^{-\frac{i\alpha }{\sqrt{2}}}{e}^{-\frac{\alpha }{\sqrt{2}}}$$$$\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{\frac{1}{4i}{e}^{-\frac{\alpha }{\sqrt{2}}}{e}^{\frac{i\alpha }{\sqrt{2}}}-\frac{1}{4i}{e}^{-\frac{\alpha }{\sqrt{2}}}{e}^{-\frac{i\alpha }{\sqrt{2}}}\}$$$$=\frac{\pi }{2}{e}^{-\frac{\alpha }{\sqrt{2}}}\left\{2isin\left(\frac{\alpha }{\sqrt{2}}\right)\right\}=i\pi e^{-\frac{\alpha }{\sqrt{2}}}sin\left(\frac{\alpha }{\sqrt{2}}\right)\Rightarrow$$$$2A=\pi e^{-\frac{\alpha }{\sqrt{2}}}sin\left(\frac{\alpha }{\sqrt{2}}\right)\Rightarrow A=\frac{\pi }{2}{e}^{-\frac{\alpha }{\sqrt{2}}}sin\left(\frac{\alpha }{\sqrt{2}}\right).$$$$$$$$$$

Commented by mathmax by abdo last updated on 03/Oct/19

$$nevermindsir.$$

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