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Question Number 184866 by paul2222 last updated on 12/Jan/23

Commented by Frix last updated on 13/Jan/23

$$\mathrm{Waiting}...$$$$\mathit{B}\mathit{I}\mathit{G}\mathit{G}\mathit{O}\mathit{D}\mathit{O}\mathit{T}$$

Answered by Mathspace last updated on 13/Jan/23

$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin(x-\frac{1}{x})}{x+\frac{1}{x}}dx$$$$I={\int }_{-\mathrm{\infty }}^{-\mathrm{\infty }}\frac{sin\left(\frac{x^2-1}{x}\right)}{\frac{x^2+1}{x}}dx$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{xsin\left(\frac{x^2-1}{x}\right)}{x^2+1}dx$$$$=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{xe}^{i\left(\frac{x^2-1}{x}\right)}}{x^2+1}dx\right)$$$$letf\left(z\right)=\frac{{ze}^{i\left(\frac{z^2-1}{z}\right)}}{z^2+1}=\frac{{ze}^{i\left(\frac{z^2-1}{z}\right)}}{(z-i)(z+i)}$$$$the\rho olesof\phi areiand-i$$$$residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi Res(f,i)$$$$=2i\pi .\frac{{ie}^{i\left(\frac{i^2-1}{i}\right)}}{2i}=i\pi e^{-2}=\frac{i\pi }{e^2}\Rightarrow$$$$I=\frac{\pi }{e^2}$$

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