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Question Number 130306 by benjo_mathlover last updated on 24/Jan/21

$${\int }_0^{\mathrm{\infty }}\mathrm{x}\cos \left(\mathrm{x}\right)\ln \left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}?$$

Answered by Dwaipayan Shikari last updated on 24/Jan/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}{x}^{a}cosx{e}^{-x}dx$$$$I\left(a\right)=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{x}^a{e}^{-(1-i)x}+x^a{e}^{-(1+i)a}dx$$$$=\frac{1}{2}\mathrm{\Gamma }(a+1)(\frac{1}{{(1-i)}^{a+1}}+\frac{1}{{(1+i)}^{a+1}})=\frac{\mathrm{\Gamma }(a+1)}{2^{\frac{a+1}{2}}}cos\left(\frac{\pi }{4}(a+1)\right)$$$$I^{\prime }\left(a\right)=\frac{{\mathrm{\Gamma }}^{\prime }(a+1)}{2^{\frac{a+1}{2}}}cos\left(\frac{\pi }{4}(a+1)\right)-\frac{\pi }{4}\frac{\mathrm{\Gamma }(a+1)}{2^{\frac{a+1}{2}}}sin\left(\frac{\pi }{4}(a+1)\right)-\frac{1}{2}.\frac{\mathrm{\Gamma }(a+1)}{2^{\frac{a}{2}+\frac{1}{2}}}log\left(2\right)cos\left(\frac{\pi }{4}(a+1)\right)$$$$I^{\prime }\left(1\right)=-\frac{\pi }{4}.\frac{1}{2}=-\frac{\pi }{8}$$

Answered by mnjuly1970 last updated on 24/Jan/21

$$\mathrm{\Omega }=\mathcal{R}e{\int }_0^{\mathrm{\infty }}{xe}^{-ix}ln\left(x\right)e^{-x}dx$$$$=\mathcal{R}e{\int }_0^{\mathrm{\infty }}{xe}^{-x(1+i)}ln\left(x\right)dx$$$$=\mathcal{R}e\left(\frac{d}{ds}{\int }_0^{\mathrm{\infty }}{x}^s{e}^{-x(1+i)}dx\right){\mid }_{s=1}$$$$=\mathcal{R}e\left(\frac{d}{ds}\left[{\int }_0^{\mathrm{\infty }}{x}^s{e}^{-x(1+i)}dx\right]\right){\mid }_{s=1}$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}{x}^{1+s}{e}^{-x(1+i)}dx\stackrel{x(1+i)=t}{=}{\int }_0^{\mathrm{\infty }}\frac{t^s}{{(1+i)}^{s+1}}{e}^{-t}dt$$$$\mathrm{\Phi }=\mathrm{\Gamma }(s+1){(1+i)}^{-s-1}$$$$\text{Missing left or extra right}$$$$\frac{d\mathrm{\Phi }}{ds}{\mid }_{s=1}={\mathrm{\Gamma }}^{\prime }\left(2\right){(1+i)}^{-2}+(-ln(1+i))e^{-2ln(1+i)}\mathrm{\Gamma }\left(2\right)$$$$=\psi \left(2\right){\left(2i\right)}^{-1}+(-ln(1+i))e^{-ln\left(2i\right)}.1$$$$=[-\psi \left(2\right)\ast \frac{i}{2}=Imaginary]-ln\left(\sqrt{2}{e}^{\frac{i\pi }{4}}\right)\ast \frac{1}{2i}$$$$\therefore \mathrm{\Omega }=-\frac{\pi }{8}...✓✓$$$$$$$$$$$$$$

Commented by benjo_mathlover last updated on 24/Jan/21

$$\mathrm{the}\mathrm{ans}\mathrm{is}-\frac{\pi }{8}$$

Commented by mnjuly1970 last updated on 24/Jan/21

$$thanksalot...$$$$$$

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