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Question Number 96198 by mathmax by abdo last updated on 30/May/20

$$\mathrm{calculate}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{sh}\left(2\mathrm{x}\right)\right)}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}\mathrm{and}\mathrm{g}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{sh}\left(2\mathrm{x}\right)\right)}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^2}(\mathrm{a}>0)$$

Answered by mathmax by abdo last updated on 01/Jun/20

$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{sh}\left(2\mathrm{x}\right)\right)}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}\Rightarrow \mathrm{f}\left(\mathrm{a}\right){=}_{\mathrm{x}=\mathrm{at}}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{sh}\left(2\mathrm{at}\right)\right)}{{\mathrm{a}}^2({\mathrm{t}}^2+1)}\mathrm{adt}$$ $$=\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{sh}\left(2\mathrm{at}\right)\right)}{{\mathrm{t}}^2+1}\mathrm{dt}\Rightarrow 2\mathrm{f}\left(\mathrm{a}\right)=\frac{1}{\mathrm{a}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(\mathrm{sh}\left(2\mathrm{at}\right)\right)}{{\mathrm{t}}^2+1}\mathrm{dt}$$ $$\Rightarrow 2\mathrm{af}\left(\mathrm{a}\right)=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ish}\left(2\mathrm{at}\right)}}{{\mathrm{t}}^2+1}\mathrm{dt}\right)\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{ish}\left(2\mathrm{az}\right)}}{{\mathrm{z}}^2+1}\Rightarrow$$ $$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{ish}\left(2\mathrm{az}\right)}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}\mathrm{residus}\mathrm{theorem}\mathrm{give}$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})=2\mathrm{i}\pi \times \frac{{\mathrm{e}}^{\mathrm{ish}\left(2\mathrm{ai}\right)}}{2\mathrm{i}}=\pi {\mathrm{e}}^{\mathrm{ish}\left(2\mathrm{ai}\right)}\mathrm{but}$$ $$\mathrm{sh}\left(2\mathrm{ai}\right)=\sin \left(2\mathrm{a}\right)\Rightarrow {\mathrm{e}}^{\mathrm{ish}\left(2\mathrm{ai}\right)}={\mathrm{e}}^{\mathrm{isin}\left(2\mathrm{a}\right)}=\cos \left(\sin \left(2\mathrm{a}\right)\right)+\mathrm{isin}\left(\sin \left(2\mathrm{a}\right)\right)\Rightarrow$$ $$2\mathrm{af}\left(\mathrm{a}\right)=\mathrm{Re}\left({\int }_0^{\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}\right)=\cos \left(\sin \left(2\mathrm{a}\right)\right)\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{\cos \left(\sin \left(2\mathrm{a}\right)\right)}{2\mathrm{a}}$$

Commented bymathmax by abdo last updated on 01/Jun/20

$$\mathrm{we}\mathrm{have}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{a}\mathrm{coz}\left(\mathrm{sh}\left(2\mathrm{x}\right)\right)}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^2}\mathrm{dx}\Rightarrow$$ $${\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{sh}\left(2\mathrm{x}\right)\right)}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^2}\mathrm{dx}=-\frac{{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)}{2\mathrm{a}}\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)=\frac{\cos \left(\sin \left(2\mathrm{a}\right)\right)}{2\mathrm{a}}\Rightarrow$$ $${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{-2\cos \left(2\mathrm{a}\right)\sin (\sin \left(2\mathrm{a}\right)\left(2\mathrm{a}\right)-2\cos (\sin \left(2\mathrm{a}\right)}{{4\mathrm{a}}^2}\Rightarrow$$ $${\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{sh}\left(2\mathrm{x}\right)\right)}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^2}\mathrm{dx}=\frac{2\mathrm{acos}\left(2\mathrm{a}\right)\sin (\sin \left(2\mathrm{a}\right)+\cos \left(\sin \left(2\mathrm{a}\right)\right)}{{4\mathrm{a}}^3}$$

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