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Question Number 132877 by mnjuly1970 last updated on 17/Feb/21

$$....nicecalculus...$$$$provethat:$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right).{log}^2\left(x\right)}{x}$$$$=\frac{\pi }{24}(12{\gamma }^2+{\pi }^2).....$$

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

$$\varphi \left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x^{\alpha }}dx$$$$\varphi \left(\alpha \right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{t}^{\alpha -1}{e}^{-xt}sin\left(x\right)dtdx$$$$=\frac{1}{2i\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{t}^{\alpha -1}{e}^{-(t-i)x}-t^{\alpha -1}{e}^{-(t+i)x}dx$$$$=\frac{1}{2i\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^{\mathrm{\infty }}\frac{t^{\alpha -1}}{t-i}-\frac{t^{\alpha -1}}{t+i}dt$$$$=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^{\mathrm{\infty }}\frac{t^{\alpha -1}}{t^2+1}dt=\frac{1}{2\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{\alpha }{2}-1}}{{(u+1)}^{1-\frac{\alpha }{2}+\frac{\alpha }{2}}}du$$$$=\frac{1}{2\mathrm{\Gamma }\left(\alpha \right)}.\frac{\mathrm{\Gamma }\left(\frac{\alpha }{2}\right)\mathrm{\Gamma }(1-\frac{\alpha }{2})}{\mathrm{\Gamma }\left(1\right)}=\frac{\pi }{2\mathrm{\Gamma }\left(\alpha \right)sin\left(\frac{\pi \alpha }{2}\right)}$$$${\varphi }^{″}\left(\alpha \right){\mid }_{\alpha =1}={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right){log}^2\left(x\right)}{x}dx=\frac{\pi }{24}(12{\gamma }^2+{\pi }^2)$$

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

$${\varphi }^{\prime }\left(\alpha \right)=-(\frac{{\pi }^2}{4\mathrm{\Gamma }\left(\alpha \right)}.\frac{cos\left(\frac{\pi }{2}\alpha \right)}{{sin}^2\left(\frac{\pi }{2}\alpha \right)}+\frac{\frac{{\pi }^2}{2}\mathrm{\Gamma }\left(\alpha \right).\frac{cos\left(\frac{\pi }{2}\alpha \right)}{{sin}^2\left(\frac{\pi }{2}\alpha \right)}}{2{\mathrm{\Gamma }}^2\left(\alpha \right)}+\frac{\pi {\mathrm{\Gamma }}^{\prime }\left(\alpha \right)cosec\left(\frac{\pi }{2}2\alpha \right)}{2{\mathrm{\Gamma }}^2\left(\alpha \right)})$$$${\varphi }^{\prime }\left(1\right)=\frac{\pi \gamma }{2}$$

Commented by mnjuly1970 last updated on 17/Feb/21

$$withthankingsirpayan...$$$$$$

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