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Question Number 138860 by qaz last updated on 19/Apr/21

$$Prove::{\int }_0^{\mathrm{\infty }}\frac{x^3-\sin {}^{3}x}{x^5}dx=\frac{13}{32}\pi$$

Answered by Dwaipayan Shikari last updated on 19/Apr/21

$${\int }_0^{\mathrm{\infty }}\frac{x^3-{sin}^{3}x}{x^5}dx$$$${\int }_0^{\mathrm{\infty }}\frac{x^3-(3sinx-sin3x)/4}{x^5}dx$$$$J\left(a\right)+G\left(3\right)=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{4x^3-3sin\left(ax\right)}{x^5}+\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{sin3x}{x^5}dx$$$$J^{″}\left(a\right)=\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{sin\left(ax\right)}{x^3}dx=\frac{3a^2}{4}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t^3}dt=\frac{3a^2}{4}.\frac{\pi }{2\mathrm{\Gamma }\left(3\right)sin\left(\frac{3\pi }{2}\right)}$$$$=-\frac{3a^2\pi }{16}\Rightarrow J\left(a\right)=-\frac{a^4\pi }{64}+Ka+C$$$$K=0C=0$$$$G\left(3\right)=\frac{3^4}{4}.\frac{\pi }{2\mathrm{\Gamma }\left(5\right)sin\left(\frac{5\pi }{2}\right)}=\frac{81\pi }{192}=\frac{27\pi }{64}$$$$J^{″}\left(1\right)+G\left(3\right)=\frac{-\pi }{64}+\frac{27\pi }{64}=\frac{13\pi }{32}$$

Commented by qaz last updated on 19/Apr/21

$$thankyousir$$

Commented by mathmax by abdo last updated on 20/Apr/21

$$\mathrm{what}\mathrm{mean}\mathrm{J}\left(\mathrm{a}\right)\mathrm{sir}...$$

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