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Question Number 177665 by Ahmed777hamouda last updated on 07/Oct/22

Answered by Ar Brandon last updated on 07/Oct/22

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sin x^{2}dx={\int }_0^{\mathrm{\infty }}\frac{\sin t}{\sqrt{t}}dt=\frac{\pi }{2\mathrm{\Gamma }\left(\frac{1}{2}\right)\sin (\frac{\pi }{2}\cdot \frac{1}{2})}=\sqrt{\frac{\pi }{2}}$$

Answered by Mathspace last updated on 07/Oct/22

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}cos\left(x^2\right)dx-i{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}sin\left(x^2\right)dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{ix}^2}dx(\sqrt{i}x=t)$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-t^2}\frac{dt}{\sqrt{i}}=\frac{1}{\sqrt{i}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-t^2}dt$$$$=e^{-\frac{i\pi }{4}}\sqrt{\pi }=\sqrt{\pi }\{cos\left(\frac{\pi }{4}\right)-isin\left(\frac{\pi }{4}\right)\}$$$$=\sqrt{\pi }\{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\}$$$$\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}cos\left(x^2\right)dx=\sqrt{\frac{\pi }{2}}$$$$and{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}sin\left(x^2\right)dx=\sqrt{\frac{\pi }{2}}$$$$anotherwaywithresidus$$$$theorembutverylong$$

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