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Question Number 33978 by abdo imad last updated on 28/Apr/18

$$letf\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{sin\left(x^2\right)e^{-{tx}^2}}{x^2}dxwitht>0$$$$findasimpleformof{f}^{{}^{\prime }}\left(t\right).$$

Commented by math khazana by abdo last updated on 01/May/18

$$afterverifyingconditionofderivabityforfwehave$$$$f^{{}^{\prime }}\left(t\right)={\int }_0^{\mathrm{\infty }}-x^2\frac{sin\left(x^2\right)e^{-{tx}^2}}{x^2}dx$$$$=-{\int }_0^{\mathrm{\infty }}sin\left(x^2\right)e^{-{tx}^2}dx\Rightarrow 2f^{{}^{\prime }}\left(x\right)=-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}sin\left(x^2\right)e^{-{tx}^2}dx$$$$=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{ix}^2}{e}^{-{tx}^2}dx\right)but$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{ix}^2}{e}^{-{tx}^2}dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-(t+i)x^2}dx$$$${=}_{\sqrt{t+i}x=u}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{t+i}}=\frac{\sqrt{\pi }}{\sqrt{t+i}}but$$$$\mid t+i\mid =\sqrt{1+t^2}\Rightarrow t+i=\sqrt{1+t^2}(\frac{t}{\sqrt{1+t^2}}+\frac{i}{\sqrt{1+t^2}})$$$$=r{e}^{i\theta }\Rightarrow r=\sqrt{1+t^2}andcos\theta =\frac{t}{\sqrt{1+t^2}}and$$$$sin\theta =\frac{1}{\sqrt{1+t^2}}\Rightarrow tan\theta =\frac{1}{t}\Rightarrow \theta =arctan\left(\frac{1}{t}\right)$$$$t+i={(1+t^2)}^{\frac{1}{2}}{e}^{i(\frac{\pi }{2}-arctant)}\Rightarrow$$$$\sqrt{t+i}={(1+t^2)}^{\frac{1}{4}}{e}^{i(\frac{\pi }{4}-\frac{1}{2}arctant)}$$$$\frac{1}{\sqrt{t+i}}={(1+t^2)}^{-\frac{1}{4}}{e}^{i(\frac{1}{2}arctant-\frac{\pi }{4})}\Rightarrow$$$$2f^{{}^{\prime }}\left(x\right)={(1+t^2)}^{-\frac{1}{4}}sin(\frac{1}{2}arctant-\frac{\pi }{4})\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{1}{2}{(1+t^2)}^{-\frac{1}{4}}sin(\frac{1}{2}arctant-\frac{\pi }{4}).$$

Commented by math khazana by abdo last updated on 01/May/18

$$f^{{}^{\prime }}\left(t\right)=\frac{1}{2}{(1+t^2)}^{-\frac{1}{4}}sin(\frac{1}{2}arctant-\frac{\pi }{4}).$$

Answered by candre last updated on 30/Apr/18

$$f\left(t\right)=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\sin x^2{e}^{-{tx}^2}}{x^2}dx$$$$letsusethefactthat$$$$\underset{-\mathrm{\infty }}{\stackrel{+\mathrm{\infty }}{\int }}{e}^{-{ax}^2}dx=\sqrt{\frac{\pi }{a}}$$$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$$$weget$$$$\frac{df}{dt}=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\partial }{\partial t}\frac{\sin x^2{e}^{-{tx}^2}}{x^2}dx$$$$=-\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\sin x^2{e}^{-{tx}^2}dx$$$$=-\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{e^{{ix}^2}-e^{-{ix}^2}}{2i}{e}^{-{tx}^2}dx$$$$=-\frac{1}{2i}[\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{e}^{(i-t)x^2}dx-\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{e}^{-(i+t)x^2}dx]$$$$=\frac{\sqrt{\pi }}{4i}(\sqrt{\frac{1}{t+i}}-\sqrt{\frac{1}{t-i}})$$

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