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Question Number 132987 by mnjuly1970 last updated on 17/Feb/21

$$....mathematicalanalysis...$$$$provethat::\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{{sin}^3\left(x\right)}{x^3}dx=\frac{3\pi }{8}$$$$\ast \ast \ast \ast ..........$$$$$$

Answered by Dwaipayan Shikari last updated on 17/Feb/21

$${\int }_0^{\mathrm{\infty }}\frac{{sin}^{3}x}{x^3}dx$$$$=\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{sinx}{x^3}-\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{sin3x}{x^3}dx$$$$=\frac{3}{4}.\frac{\pi }{4sin\left(\frac{3\pi }{2}\right)}-\frac{9}{4}{\int }_0^{\mathrm{\infty }}\frac{sinu}{u^3}du{\int }_0^{\mathrm{\infty }}\frac{sinx}{x^{\mu }}du=\frac{\pi }{2\mathrm{\Gamma }\left(\mu \right)sin\left(\frac{\pi }{2}\mu \right)}$$$$=-\frac{9}{4.4sin\left(\frac{3\pi }{2}\right)}-\frac{3\pi }{16}=\frac{9\pi }{16}-\frac{3\pi }{16}=\frac{3\pi }{8}$$

Commented by mnjuly1970 last updated on 18/Feb/21

$$thanksalotmrpower...$$

Commented by Dwaipayan Shikari last updated on 18/Feb/21

$$Mrpower??$$$$hahaaa....sir$$

Commented by mnjuly1970 last updated on 18/Feb/21

$$sorrysorry...$$$$masterDwaipayan..$$$$$$

Answered by mnjuly1970 last updated on 18/Feb/21

$$anothersolution...$$$$\phi \left(a\right)={\int }_0^{\mathrm{\infty }}\frac{{sin}^3\left(ax\right)}{x^3}dx$$$$\mathit{\varphi }=\phi \left(1\right)...$$$${\phi }^{\prime }\left(a\right)=\frac{\partial }{\partial a}\left({\int }_0^{\mathrm{\infty }}\frac{{sin}^3\left(ax\right)}{x^3}dx\right)$$$$={\int }_0^{\mathrm{\infty }}\frac{3xcos\left(ax\right){sin}^2\left(ax\right)}{x^3}dx$$$$=3{\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(ax\right)cos\left(ax\right)}{x^2}dx$$$$=\frac{3}{2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(2ax\right)sin\left(ax\right)}{x^2}dx$$$$=\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{cos\left(ax\right)-cos\left(3ax\right)}{x^2}dx$$$${\phi }^{″}\left(a\right)=\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{-sin\left(ax\right)}{x}dx+\frac{9}{4}{\int }_0^{\mathrm{\infty }}\frac{sin\left(3ax\right)}{x}dx$$$$=\frac{3\pi }{4}\Rightarrow {\phi }^{\prime }\left(a\right)=\frac{3}{4}\pi a+C$$$${\phi }^{\prime }\left(0\right)=0=C$$$${\phi }^{\prime }\left(a\right)=\frac{3}{4}\pi a\Rightarrow \phi \left(a\right)=\frac{3}{8}\pi a^2+K$$$$\phi \left(0\right)=0=K$$$$\phi \left(a\right)=\frac{3}{8}\pi a^2\Rightarrow \mathit{\varphi }=\phi \left(1\right)=\frac{3}{8}\pi ✓✓$$

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