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Question Number 127432 by mnjuly1970 last updated on 29/Dec/20

$$...\mathrm{CALCULUS}...$$$$$$$$\varnothing ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}cos\left(\frac{\pi x^2}{2}\right)dx=?$$$$$$

Answered by Dwaipayan Shikari last updated on 29/Dec/20

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sqrt{\frac{2}{\pi }}cos\left(u^2\right)du\sqrt{\frac{\pi }{2}}x=u\Rightarrow$$$$=\sqrt{\frac{8}{\pi }}{\int }_0^{\mathrm{\infty }}cos\left(u^2\right)du=1$$$${\int }_0^{\mathrm{\infty }}cos\left(u^2\right)du=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{u^{2}i}+e^{-u^{2}i}du=\frac{1}{4}{\int }_0^{\mathrm{\infty }}{t}^{-\frac{1}{2}}{e}^{it}+t^{-\frac{1}{2}}{e}^{-it}dt$$$$=\frac{{\left(i\right)}^{-\frac{1}{2}}i}{4}{\int }_0^{\mathrm{\infty }}{m}^{-\frac{1}{2}}{e}^{-m}dm-\frac{{(-i)}^{-\frac{1}{2}}i}{4}{\int }_0^{\mathrm{\infty }}{j}^{-\frac{1}{2}}{e}^{-j}djt=-ijt=im$$$$=2i\mathrm{\Gamma }\left(\frac{1}{2}\right)\left(\frac{e^{-\frac{\pi }{4}i}-e^{\frac{\pi }{4}i}}{4}\right)=2\sqrt{\frac{\pi }{16}}sin\left(\frac{\pi }{4}\right)=\sqrt{\frac{\pi }{8}}$$

Commented by mnjuly1970 last updated on 29/Dec/20

$$thankyoumaster...$$$$$$

Commented by Dwaipayan Shikari last updated on 31/Dec/20

$$Haveagreatyearsir!$$

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