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Question Number 130949 by mnjuly1970 last updated on 31/Jan/21

$$...nicecalculus...$$$$calculate:$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{S}i\left(x\right)}{x^{\frac{3}{2}}}dx\stackrel{???}{=}\sqrt{8\pi }$$$$\mathrm{S}i\left(x\right)={\int }_0^x\frac{sin\left(t\right)}{t}dt$$

Answered by mindispower last updated on 31/Jan/21

$$\int \frac{si\left(x\right)}{x^{\frac{3}{2}}}dx=-\frac{2si\left(x\right)}{\sqrt{x}}+2\int \frac{sin\left(x\right)}{\sqrt{x}.x}dx$$$$putx=u^2,in2nd$$$$2\int \frac{sin\left(x\right)}{x\sqrt{x}}dx=4\int \frac{sin\left(u^2\right)}{u^2}du$$$$bypart=-\frac{4sin\left(u^2\right)}{u}+8\int cos\left(u^2\right)du...E$$$${\int }_0^{t}cos\left(\frac{\pi }{2}{x}^2\right)dx=C\left(t\right)fersnelintegral$$$$u=\sqrt{\frac{\pi }{2}}y$$$$\int \sqrt{\frac{\pi }{2}}ycos\left(\frac{\pi y^2}{2}\right)dy=\sqrt{\frac{\pi }{2}}C\left(y\right)$$$$weget$$$$E=-\frac{4sin\left(u^2\right)}{u}+4\sqrt{2\pi }C\left(\sqrt{\frac{2}{\pi }}u\right)$$$$\int \frac{si\left(x\right)}{x^{\frac{3}{2}}}dx=\frac{-2si\left(x\right)}{\sqrt{x}}-\frac{4sin\left(x\right)}{\sqrt{x}}+4\sqrt{2\pi }.C(\sqrt{\frac{2}{\pi }}.\sqrt{x})$$$$x\to \mathrm{\infty }$$$$\underset{x\to \mathrm{\infty }}{\lim }C\left(x\right)=\frac{1}{2}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{si\left(x\right)}{x\sqrt{x}}dx=\underset{x\to \mathrm{\infty }}{\lim }(-\frac{2si\left(x\right)}{\sqrt{x}}-\frac{4sin\left(x\right)}{\sqrt{x}}+4\sqrt{2\pi }Ci\left(\sqrt{\frac{2x}{\pi }}\right))$$$$=2\sqrt{2\pi }=\sqrt{8\pi }$$$$$$

Commented by mnjuly1970 last updated on 31/Jan/21

$$veryniceasalways$$$$sirmindispower...$$$$Allahkeepyou..$$

Commented by mindispower last updated on 31/Jan/21

$$inshallahalawayswithepleasur$$

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