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Question Number 115507 by mnjuly1970 last updated on 26/Sep/20

$$.......matematicalanalysis...$$ $$$$ $$provethat:::$$ $$a>0::\left[\begin{array}{c}i:{\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(ax\right)}{x^{\frac{3}{2}}}dx=\sqrt{\pi a}\\ ii:{\int }_0^{\mathrm{\infty }}\frac{{sin}^3\left(ax\right)}{\sqrt{x}}dx=\frac{-1+3\sqrt{3}}{4}\sqrt{\frac{\pi }{6a}}\end{array}\right]$$ $$$$ $$...m.n.july.1970...$$ $$$$

Answered by mathdave last updated on 26/Sep/20

$$\left(1\right)letI={\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\left(ax\right)}{x^{\frac{3}{2}}}dx={\int }_0^{\mathrm{\infty }}\frac{\sin \left(2ax\right)}{x^{\frac{3}{2}}}=\frac{\pi a^{n-1}}{2\mathrm{\Gamma }\left(n\right)\sin \left(\frac{\pi n}{2}\right)}\left(callmazidentity\right)$$ $$I=\frac{\pi {\left(2a\right)}^{\frac{1}{2}}}{2.\frac{1}{2}\sqrt{\pi .}\sin \left(\frac{3\pi }{4}\right)}=\frac{\sqrt{a\pi }.\sqrt{2}}{\sin \left(\frac{\pi }{4}\right)}=\frac{\sqrt{a\pi }.\sqrt{2}}{\frac{1}{\sqrt{2}}}=2\sqrt{a\pi }$$ $$\because {\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\left(ax\right)}{x^{\frac{3}{2}}}dx=2\sqrt{a\pi }Q.E.D$$ $$\left(2\right)letI={\int }_0^{\mathrm{\infty }}\frac{{\sin }^3\left(ax\right)}{\sqrt{x}}dx$$ $$but\sin \left(3x\right)=3\sin x-{4\sin }^{3}x,{\sin }^{3}x=\frac{3\sin x-\sin \left(3x\right)}{4}$$ $$I=\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{\sin \left(ax\right)}{x^{\frac{1}{2}}}dx-\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{\sin \left(3ax\right)}{x^{\frac{1}{2}}}dx$$ $$I=\frac{3}{4}.\frac{\pi a^{-\frac{1}{2}}}{2\sqrt{\pi }.\sin \left(\frac{\pi }{4}\right)}-\frac{1}{4}.\frac{\pi {\left(3a\right)}^{-\frac{1}{2}}}{2\sqrt{\pi }.\sin \left(\frac{\pi }{4}\right)}=\frac{3}{4}.\frac{\sqrt{\pi }.\sqrt{2}}{2\sqrt{a}}-\frac{1}{4}.\frac{\sqrt{\pi }.\sqrt{2}}{2\sqrt{3a}}$$ $$I=\frac{3\sqrt{\pi }}{4\sqrt{2a}}-\frac{\sqrt{\pi }}{4\sqrt{6a}}=\frac{3\sqrt{3}.\sqrt{\pi }-\sqrt{\pi }}{4\sqrt{6a}}=\frac{(-1+3\sqrt{3})\sqrt{\pi }}{4\sqrt{6a}}=\frac{-1+3\sqrt{3}}{4}\sqrt{\frac{\pi }{6a}}?$$ $$\because {\int }_0^{\mathrm{\infty }}\frac{{\sin }^3\left(ax\right)}{\sqrt{x}}dx=\frac{-1+3\sqrt{3}}{4}\sqrt{\frac{\pi }{6a}}Q.E.D$$ $$bymathdave\left(26/09/2020\right)$$

Commented bymnjuly1970 last updated on 26/Sep/20

$$thankyou$$ $$mazidentity$$ $$i:{\int }_0^{\mathrm{\infty }}\frac{cos\left(x\right)}{x^p}dx=\frac{\pi }{2\mathrm{\Gamma }\left(p\right)cos\left(\frac{p\pi }{2}\right)}0<p<1$$ $$ii:{\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x^p}dx=\frac{\pi }{2\mathrm{\Gamma }\left(p\right)sin\left(\frac{p\pi }{2}\right)}0<p⩽1$$ $$murrayspiegeladvanced$$ $$calculus.....$$ $$$$

Commented bymnjuly1970 last updated on 26/Sep/20

$$thanksalotmrdave$$

Answered by Bird last updated on 28/Sep/20

$$I={\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(ax\right)}{x^{\frac{3}{2}}}dx$$ $$\Rightarrow I{=}_{ax=t}{\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}t}{{\left(\frac{t}{a}\right)}^{\frac{3}{2}}}\times \frac{dt}{a}$$ $$=a^{\frac{3}{2}-1}{\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}t}{t^{\frac{3}{2}}}dt$$ $$==\sqrt{a}{\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}t}{t^{\frac{3}{2}}}dtbyparts$$ $${\int }_0^{\mathrm{\infty }}{t}^{-\frac{3}{2}}{sin}^{2}tdt$$ $$={\left[\frac{1}{1-\frac{3}{2}}{t}^{1-\frac{3}{2}}{sin}^{2}t\right]}_0^{\mathrm{\infty }}$$ $$-{\int }_0^{\mathrm{\infty }}-2t^{-\frac{1}{2}}\times 2sintcostdt$$ $$=2{\int }_0^{\mathrm{\infty }}\frac{sin\left(2t\right)}{\sqrt{t}}dt$$ $${=}_{\sqrt{t}=u}2{\int }_0^{\mathrm{\infty }}\frac{sin\left(2u^2\right)}{u}\left(2u\right)du$$ $$=4{\int }_0^{\mathrm{\infty }}sin\left(2u^2\right)du$$ $$=-4Im\left({\int }_0^{\mathrm{\infty }}{e}^{-2{iu}^2}du\right)$$ $${\int }_0^{\mathrm{\infty }}{e}^{-{\left(\sqrt{2i}u\right)}^2}du{=}_{\sqrt{2i}u=z}{\int }_0^{\mathrm{\infty }}{e}^{-z^2}\frac{dz}{\sqrt{2i}}$$ $$=\frac{1}{\sqrt{2}}{e}^{-\frac{i\pi }{4}}\times \frac{\sqrt{\pi }}{2}$$ $$=\frac{\sqrt{\pi }}{2\sqrt{2}}(\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2})$$ $$=\frac{\sqrt{\pi }}{4}-\frac{i\sqrt{\pi }}{4}\Rightarrow$$ $${\int }_0^{\mathrm{\infty }}{t}^{-\frac{3}{2}}{sin}^{2}tdt=-4(-\frac{\sqrt{\pi }}{4})=\sqrt{\pi }$$ $$\Rightarrow I=\sqrt{a}\times \sqrt{\pi }=\sqrt{\pi a}$$ $$$$

Commented bymnjuly1970 last updated on 28/Sep/20

$$thankyousomuchsirmath$$ $$bird..$$

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