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Question Number 130432 by mnjuly1970 last updated on 25/Jan/21

$$...nicecalculus...$$$$calculate::$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}xln\left(x\right)e^{-x}sin\left(x\right)dx=?$$$$$$

Answered by Dwaipayan Shikari last updated on 25/Jan/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}{x}^a{e}^{-x}sin\left(x\right)dx=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}{x}^a{e}^{-x(1-i)}-x^a{e}^{-x(1+i)}dx$$$$=\frac{\mathrm{\Gamma }(a+1)}{2i{(1-i)}^{a+1}}-\frac{\mathrm{\Gamma }(a+1)}{2i{(1+i)}^{a+1}}=\frac{\mathrm{\Gamma }(a+1)}{2^{\frac{a+1}{2}}}sin\left(\frac{\pi }{4}(a+1)\right)$$$$I^{\prime }\left(a\right)=\frac{{\mathrm{\Gamma }}^{\prime }(a+1)}{2^{\frac{a+1}{2}}}sin\left(\frac{\pi }{4}(a+1)\right)+\frac{\pi }{4}.\frac{\mathrm{\Gamma }(a+1)}{2^{\frac{a+1}{2}}}cos\left(\frac{\pi }{4}(a+1)\right)-\frac{log\left(2\right)}{2}.\frac{\mathrm{\Gamma }(a+1)}{2^{\frac{a+1}{2}}}sin\left(\frac{\pi }{4}(a+1)\right)$$$$I^{\prime }\left(1\right)=\frac{1-\gamma }{2}-\frac{log\left(2\right)}{4}=\frac{1}{2}-\frac{1}{4}(2\gamma +log\left(2\right))$$

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