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Question Number 61528 by maxmathsup by imad last updated on 04/Jun/19

$$find{\int }_0^{\mathrm{\infty }}cos\left({zx}^2\right)dxwithz\in C.$$

Commented by maxmathsup by imad last updated on 04/Jun/19

$$letI={\int }_0^{\mathrm{\infty }}cos\left({zx}^2\right)dx\Rightarrow 2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}cos\left({zx}^2\right)letz=a+ib\Rightarrow$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}cos\left((a+ib)x^2\right)butcos\left(z\right)=ch\left(iz\right)\Rightarrow$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}ch\left(i(a+ib)x^2\right)dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}ch({iax}^2-{bx}^2)dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i({iax}^2-{bx}^2)}+e^{-i({iax}^2-{bx}^2)}}{2}dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{-{ax}^2-{ibx}^2}+e^{{ax}^2+{ibx}^2}}{2}dx=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-(a+ib)x^2}dx+\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-(-a-ib)x^2}dx$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-(a+ib)x^2}dx{=}_{\sqrt{a+ib}x=t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-t^2}\frac{dt}{\sqrt{a+ib}}=\frac{\pi }{\sqrt{a+ib}}=\frac{\pi }{\sqrt{z}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-(-a-ib)x^2}dx{=}_{\sqrt{-a-ib}x=t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-t^2}\frac{dt}{\sqrt{-a-ib}}=\frac{\pi }{\sqrt{-a-ib}}$$$$=\frac{\pi }{\sqrt{-1}\sqrt{a+ib}}=\frac{\pi }{i\sqrt{z}}\Rightarrow 2I=\frac{\pi }{2\sqrt{z}}+\frac{\pi }{2i\sqrt{z}}=\frac{\pi }{2\sqrt{z}}-\frac{i\pi }{2\sqrt{z}}\Rightarrow I=\frac{\pi }{4}\frac{1-i}{\sqrt{z}}$$$$ifwetakez={re}^{i\theta }wegetI=\frac{\pi }{4}\frac{1-i}{\sqrt{r}{e}^{i\frac{\theta }{2}}}=\frac{\pi }{4\sqrt{r}}\sqrt{2}{e}^{-\frac{i\pi }{4}}{e}^{-\frac{i\theta }{2}}=\frac{\pi \sqrt{2}}{4\sqrt{r}}{e}^{i(-\frac{\pi }{4}-\frac{\theta }{2})}$$$$I=\frac{\pi \sqrt{2}}{4\sqrt{r}}{e}^{-i(\frac{\pi }{4}+\frac{\theta }{2})}.$$

Commented by Smail last updated on 05/Jun/19

$$Youmean{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-t^2}dt=\sqrt{\pi }not\pi$$

Commented by maxmathsup by imad last updated on 08/Jun/19

$$yessir....$$

Commented by maxmathsup by imad last updated on 08/Jun/19

$$dueto{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-t^2}dt=\sqrt{\pi }thefinalanswerisI=\frac{\sqrt{2\pi }}{4\sqrt{r}}{e}^{-i(\frac{\pi }{4}+\frac{\theta }{2})}$$$$$$

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