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Question Number 41678 by math khazana by abdo last updated on 11/Aug/18

$$provethat{\int }_0^{\mathrm{\infty }}cos\left(x^2\right)dx={\int }_0^{\mathrm{\infty }}sin\left(x^2\right)dxbyusing$$$$onlyseries.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18

$$p+iq={\int }_0^{\mathrm{\infty }}{e}^{{ix}^2}dx$$$$-t={ix}^2{i}^{2}t={ix}^2$$$$it=x^2$$$$2xdx=idt$$$$dx=\frac{idt}{2x}=\frac{idt}{2\sqrt{it}}=\frac{\sqrt{i}{t}^{\frac{-1}{2}}}{2}dt$$$${\int }_0^{\mathrm{\infty }}{e}^{-t}.\frac{\sqrt{i}{t}^{\frac{-1}{2}}}{2}dt$$$$\frac{\sqrt{i}}{2}{\int }_0^{\mathrm{\infty }}{e}^{-t}.t^{\frac{1}{2}-1}dt$$$$=\frac{\sqrt{i}}{2}\lceil \left(\frac{1}{2}\right)=\frac{\sqrt{\mathrm{\Pi }}}{2}\times \frac{\sqrt{i}}{}$$$$nowcalculatingthevalueof\sqrt{i}$$$$\sqrt{i}$$$$=\frac{1}{\sqrt{2}}\times \sqrt{1-1+2i}$$$$=\frac{1}{\sqrt{2}}\times \sqrt{1^2+i^2+2\times 1\times i}$$$$=\frac{1}{\sqrt{2}}\times \sqrt{{(1+i)}^2}$$$$=\frac{1}{\sqrt{2}}\times (1+i)$$$$=\frac{1}{\sqrt{2}}+i\times \frac{1}{\sqrt{2}}$$$$sop+iq={\int }_0^{\mathrm{\infty }}{e}^{{ix}^2}dx=\frac{\sqrt{\mathrm{\Pi }}}{2}\times \sqrt{i}=\frac{\sqrt{\mathrm{\Pi }}}{2}(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}})$$$$p=\frac{\sqrt{\mathrm{\Pi }}}{2\sqrt{2}}q=\frac{\sqrt{\mathrm{\Pi }}}{2\sqrt{2}}$$$$$$

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