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Question Number 34229 by abdo imad last updated on 03/May/18

$$calculate{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{cos\left(tx\right)}{1+x^4}dxwitht⩾0$$$$2)calculate{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^4}.$$

Commented by math khazana by abdo last updated on 04/May/18

$$letputf\left(t\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(tx\right)}{1+x^4}dx$$$$f\left(t\right)=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{itx}}{1+x^4}dx\right)letconsiderthecomplex$$$$function\phi \left(z\right)=\frac{e^{itz}}{1+z^4}wehave$$$$\phi \left(z\right)=\frac{e^{itz}}{(z^2-i)(z^2+i)}=\frac{e^{itz}}{(z-\sqrt{i})(z+\sqrt{i})(z-\sqrt{-i})(z+\sqrt{-i})}$$$$=\frac{e^{itz}}{(z-e^{i\frac{\pi }{4}})(z+e^{i\frac{\pi }{4}})(z-e^{-\frac{i\pi }{4}})(z+e^{-i\frac{\pi }{4}})}$$$$thepolesof\phi are{e}^{i\frac{\pi }{4}},-e^{i\frac{\pi }{4}},e^{-i\frac{\pi }{4}},-e^{-i\frac{\pi }{4}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (Res(\phi ,e^{i\frac{\pi }{4}})+Res(\phi ,-e^{-i\frac{\pi }{4}})$$$$letp\left(z\right)allpolesof\phi arerootsofpand$$$$p^{{}^{\prime }}\left(z\right)=4z^3\Rightarrow p^{{}^{\prime }}\left(z_k\right)=4z_k^3=\frac{4z_k^4}{z_k}=\frac{-4}{z_k}$$$$Res(\phi ,e^{i\frac{\pi }{4}})=\frac{e^{it{e}^{i\frac{\pi }{4}}}}{-4}{e}^{i\frac{\pi }{4}}=-\frac{1}{4}{e}^{i\frac{\pi }{4}}{e}^{it{e}^{i\frac{\pi }{4}}}$$$$=-\frac{1}{4}{e}^{i(\frac{\pi }{4}+t(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}))}=-\frac{1}{4}{e}^{i(\frac{\pi }{4}+\frac{t}{\sqrt{2}})-\frac{t}{\sqrt{2}}}$$$$=-\frac{1}{4}{e}^{-\frac{t}{\sqrt{2}}}{e}^{i(\frac{\pi }{4}+\frac{t}{\sqrt{2}})}$$$$Res(\phi ,-e^{-i\frac{\pi }{4}})=\frac{1}{4}{e}^{-i\frac{\pi }{4}}{e}^{-{ite}^{-i\frac{\pi }{4}}}$$$$=\frac{1}{4}{e}^{-i(\frac{\pi }{4}+t(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}))}=\frac{1}{4}{e}^{-i(\frac{\pi }{4}+\frac{t}{\sqrt{2}})-\frac{t}{\sqrt{2}}}$$$$=\frac{1}{4}{e}^{-\frac{t}{\sqrt{2}}}{e}^{-i(\frac{\pi }{4}+\frac{t}{\sqrt{2}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{-\frac{t}{\sqrt{2}}}}{4}(e^{-i(\frac{\pi }{4}+\frac{t}{\sqrt{2}})}-e^{i(\frac{\pi }{4}+\frac{t}{\sqrt{2}})})$$$$=\frac{i\pi }{2}{e}^{-\frac{t}{\sqrt{2}}}(-2isin(\frac{\pi }{4}+\frac{t}{2}))=\pi e^{-\frac{t}{\sqrt{2}}}sin(\frac{\pi }{4}+\frac{t}{2})$$$$f\left(t\right)=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz\right)\Rightarrow$$$$f\left(t\right)=\pi e^{-\frac{t}{\sqrt{2}}}sin(\frac{\pi }{4}+\frac{t}{2}).$$

Commented by math khazana by abdo last updated on 04/May/18

$$2)lettaket=0\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+x^4}=f\left(0\right)=\pi sin\left(\frac{\pi }{4}\right)$$$$=\frac{\pi }{\sqrt{2}}=\frac{\pi \sqrt{2}}{2}\Rightarrow {\int }_0^{+\mathrm{\infty }}\frac{dx}{1+x^4}=\frac{\pi \sqrt{2}}{4}.$$

Commented by math khazana by abdo last updated on 04/May/18

$$erroratthefinallines$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{i\pi }{2}{e}^{-\frac{t}{\sqrt{2}}}(-2isin(\frac{\pi }{4}+\frac{t}{\sqrt{2}}))$$$$=\pi e^{-\frac{t}{\sqrt{2}}}sin(\frac{\pi }{4}+\frac{t}{\sqrt{2}})\Rightarrow$$$$f\left(t\right)=\pi e^{-\frac{t}{\sqrt{2}}}sin(\frac{\pi }{4}+\frac{t}{\sqrt{2}}).$$

Answered by Joel578 last updated on 03/May/18

$${\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{\sin \left(\pi a\right)},0<a<1$$$$I={\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^4}(u=x^4\to du=4x^{3}dx)$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{x^{-3}}{1+u}du$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{-\frac{3}{4}}}{1+u}du=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{4}-1}}{1+u}du$$$$=\frac{1}{4}\left(\frac{\pi }{\sin \left(\pi /4\right)}\right)$$$$=\frac{\pi \sqrt{2}}{4}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/May/18

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{1+x^4}=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18

$$\frac{1}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{(1+\frac{1}{x^2})-(1-\frac{1}{x^2})}{x^2+\frac{1}{x^2}}dx$$$$\frac{1}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{(1+\frac{1}{x^2})}{{(x-\frac{1}{x})}^2+2}dx-\frac{1}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{(1-\frac{1}{x^2})}{{(x+\frac{1}{x})}^2-2}$$$$\frac{1}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{dt}_1}{t_1^2+{\left(\sqrt{2}\right)}^2}-\frac{1}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{dt}_2}{t_2^2-{\left(\sqrt{2}\right)}^2}$$$$=\frac{1}{2\sqrt{2}}\mid {tan}^{-1}\left(\frac{t_1}{\sqrt{2}}\right)\underset{0}{\stackrel{\mathrm{\infty }}{\mid }}-\frac{1}{2}.\frac{1}{2\sqrt{2}}\mid ln\mid \frac{\sqrt{2}-t_2}{\sqrt{2}+t_2}\mid$$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18

$$=\frac{1}{2\sqrt{2}}.\frac{\mathrm{\Pi }}{2}-\frac{1}{4\sqrt{2}}.ln\mid \frac{\frac{\sqrt{2}}{t_2}-1}{\frac{\sqrt{2}}{t_2}+1}\mid \to itsvalueiszero$$

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