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Question Number 131529 by mnjuly1970 last updated on 05/Feb/21

$$...advanced\ast \ast \ast \ast \ast \ast \ast \ast \ast \ast calculus...$$$$provethat:::::$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{sin\left(x^4\right)ln\left(x\right)}{x}dx=-\frac{\mathit{\pi }\mathit{\gamma }}{32}$$$$note:{\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)ln\left(x\right)}{x}dx\stackrel{why???}{=}\frac{-\mathit{\pi }\mathit{\gamma }}{2}$$$$\mathit{\varphi }\stackrel{⟨x^4=t⟩}{=}\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)ln\left(t^{\frac{1}{4}}\right)}{t^{\frac{1}{4}}}\ast \frac{1}{t^{\frac{3}{4}}}dt$$$$=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)ln\left(t\right)}{t}dt\stackrel{note}{=}\frac{-\mathit{\pi }\mathit{\gamma }}{32}$$$$(\mathit{\varphi }=-\frac{\mathit{\pi }\mathit{\gamma }}{32})$$$$notice::youwillprovethat::{\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)ln\left(x\right)}{x}dx=\frac{-\mathit{\pi }\mathit{\gamma }}{2}$$$$Hint::{\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x^{\mathit{\mu }}}dx\stackrel{???}{=}\frac{\mathit{\pi }}{2\mathbf{\Gamma }\left(\mathit{\mu }\right)sin\left(\frac{\mathit{\mu }\mathit{\pi }}{2}\right)}$$$$...m.n.july.1970...$$

Answered by Dwaipayan Shikari last updated on 06/Feb/21

$${\int }_0^{\mathrm{\infty }}\frac{sinx}{x^{\mu }}dx=\frac{1}{\mathrm{\Gamma }\left(\mu \right)}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{t}^{\mu -1}{e}^{-xt}sinxdxdt$$$$=\frac{1}{2i\mathrm{\Gamma }\left(\mu \right)}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{t}^{\mu -1}{e}^{-x(t-i)}-t^{\mu -1}{e}^{-x(t+i)}dxdt$$$$=\frac{1}{2i\mathrm{\Gamma }\left(\mu \right)}{\int }_0^{\mathrm{\infty }}\frac{t^{\mu -1}}{t-i}-\frac{t^{\mu -1}}{t+i}dt=\frac{1}{\mathrm{\Gamma }\left(\mu \right)}{\int }_0^{\mathrm{\infty }}\frac{t^{\mu -1}}{t^2+1}dt$$$$=\frac{1}{2\mathrm{\Gamma }\left(\mu \right)}{\int }_0^{\mathrm{\infty }}\frac{j^{\frac{\mu }{2}-1}}{(1+u)}dj=\frac{1}{2\mathrm{\Gamma }\left(\mu \right)}.\frac{\pi }{sin\left(\frac{\mu \pi }{2}\right)}$$$$I\left(\mu \right)=\frac{\pi }{2\mathrm{\Gamma }\left(\mu \right)sin\left(\frac{\mu \pi }{2}\right)}\Rightarrow I^{\prime }\left(\mu \right)=-\frac{{\pi }^2}{4\mathrm{\Gamma }\left(\mu \right)}cosec\left(\frac{\mu \pi }{2}\right)cot\left(\frac{\mu \pi }{2}\right)-\frac{\pi }{2sin\left(\frac{\mu \pi }{2}\right)}.\frac{{\mathrm{\Gamma }}^{\prime }\left(\mu \right)}{{\mathrm{\Gamma }}^2\left(\mu \right)}$$$$I^{\prime }\left(\mu \right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x^{\mu }}log\left(x\right)dx$$$$I^{\prime }\left(1\right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}log\left(x\right)dx=-\frac{\pi }{2}.{\mathrm{\Gamma }}^{\prime }\left(1\right)=-\frac{\pi \gamma }{2}$$$$$$

Commented by mnjuly1970 last updated on 05/Feb/21

$$taydballahmrpayan..$$

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