calculate-0-sin-3x-2-x-2-dx-

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Question Number 65922 by mathmax by abdo last updated on 05/Aug/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{sin\left(3x^2\right)}{x^2}dx$$

Commented by mathmax by abdo last updated on 08/Aug/19

$$letA={\int }_0^{\mathrm{\infty }}\frac{sin\left(3x^2\right)}{x^2}dx\Rightarrow A{=}_{\sqrt{3}x=t}3{\int }_0^{\mathrm{\infty }}\frac{sin\left(t^2\right)}{t^2}\frac{dt}{\sqrt{3}}$$$$=\frac{1}{\sqrt{3}}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t^2\right)}{t^2}dtlet\phi \left(x\right)={\int }_0^{\mathrm{\infty }}\frac{sin\left(t^2\right)e^{-{xt}^2}}{t^2}dtwithx⩾0$$$${\phi }^{{}^{\prime }}\left(x\right)=-{\int }_0^{\mathrm{\infty }}{e}^{-{xt}^2}sin\left(t^2\right)dt=-Im\left({\int }_0^{\mathrm{\infty }}{e}^{-{xt}^2+{it}^2}dt\right)and$$$${\int }_0^{\mathrm{\infty }}{e}^{(-x+i)t^2}dt{=}_{\sqrt{-x+i}t=u}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{-x+i}}$$$$=\frac{\sqrt{\pi }}{2\sqrt{-x+i}}\Rightarrow \phi \left(x\right)=\frac{\sqrt{\pi }}{2}\int \frac{dx}{\sqrt{-x+i}}+c$$$$-x+i=\sqrt{1+x^2}{e}^{iarctan(-\frac{1}{x})}=\sqrt{1+x^2}{e}^{-iarctan\left(\frac{1}{x}\right)}$$$$=\sqrt{1+x^2}{e}^{-i(\frac{\pi }{2}-arctanx)}=-i\sqrt{1+x^2}{e}^{iatctanx}\Rightarrow$$$$\sqrt{-x+i}=\sqrt{-i}{(1+x^2)}^{\frac{1}{4}}{e}^{\frac{i}{2}arctan\left(x\right)}=e^{-\frac{i\pi }{4}}{(1+x^2)}^{\frac{1}{4}}{e}^{\frac{i}{2}arctanx}\Rightarrow$$$$\phi \left(x\right)=\frac{\sqrt{\pi }}{2}{e}^{\frac{i\pi }{4}}\int {(1+x^2)}^{-\frac{1}{4}}{e}^{-\frac{i}{2}arctanx}dx+c$$$$\dots .becontinued\dots .$$

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