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Question Number 99228 by  M±th+et+s last updated on 19/Jun/20

$$$$$${\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)ln\left(x\right)}{x}dx=\frac{-\gamma \pi }{2}$$

Answered by maths mind last updated on 20/Jun/20

$$\frac{ln\left(x\right)}{x}=\frac{\partial }{\partial a}{x}^{a-1}{\mid }_{a=0}$$$${\int }_0^{+\mathrm{\infty }}sin\left(x\right)x^{a-1}dx=f\left(a\right)$$$$Wewant\underset{a\to 0}{\lim }\frac{\partial f\left(a\right)}{\partial a}$$$$f\left(a\right)=Im{\int }_0^{+\mathrm{\infty }}{e}^{ix}{x}^{a-1}dx$$$$ix=-y\Rightarrow dx=idy$$$$f\left(a\right)=Im{\int }_0^{-i\mathrm{\infty }}{e}^{-y}{i}^a{y}^{a-1}dy$$$$=Im{i}^a{\int }_0^{+\mathrm{\infty }}{e}^{-y}{y}^{a-1}dy=sin\left(\frac{\pi }{2}a\right)\mathrm{\Gamma }\left(a\right)$$$$f\left(a\right)=\frac{sin\left(\frac{\pi }{2}a\right)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(1-a)}{\mathrm{\Gamma }(1-a)}\iff f\left(a\right)=\frac{\pi sin\left(\frac{\pi }{2}a\right)}{sin\left(\pi a\right)\mathrm{\Gamma }(1-a)}...E$$$$=\frac{\pi }{2}.\frac{1}{cos\left(\frac{\pi }{2}a\right)\mathrm{\Gamma }(1-a)}$$$$f^{\prime }\left(a\right)=-\frac{\pi }{2}\left\{\frac{-\frac{\pi }{2}sin\left(\frac{\pi }{2}a\right)\mathrm{\Gamma }(1-a)-{\mathrm{\Gamma }}^{\prime }(1-a)cos\left(\frac{\pi a}{2}\right)}{{\left(cos\left(\frac{\pi a}{2}\right)\mathrm{\Gamma }(1-a)\right)}^2}\right\}$$$$f^{\prime }\left(a\right)=-\frac{\pi }{2}\frac{-{\mathrm{\Gamma }}^{\prime }\left(1\right)}{1^2}=\frac{\pi {\mathrm{\Gamma }}^{\prime }\left(1\right)}{2}=-\frac{\pi }{2}\gamma$$$$E\mathrm{\Gamma }(1-a)\mathrm{\Gamma }\left(a\right)=\frac{\pi }{sin\left(\pi a\right)}=\frac{\pi }{2}.\frac{1}{cos\left(\frac{\pi a}{2}\right)sin\left(\frac{\pi }{2}a\right)}$$$$$$

Commented by  M±th+et+s last updated on 21/Jun/20

$$verygoodprof.mathmind$$$$thankyou$$

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