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Question Number 130485 by benjo_mathlover last updated on 26/Jan/21

$$ϝ={\int }_0^{\mathrm{\infty }}{x}^5\ln \left(x\right)\cos \left(x\right)e^{-x}dx?$$

Answered by Dwaipayan Shikari last updated on 26/Jan/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}{x}^a{cosxe}^{-x}dx=\frac{\mathrm{\Gamma }(a+1)}{2^{\frac{a+1}{2}}}cos\left(\frac{\pi }{4}(a+1)\right)$$$$I^{\prime }\left(a\right)=\frac{{\mathrm{\Gamma }}^{\prime }(a+1)}{2^{\frac{a+1}{2}}}cos\left(\frac{\pi }{4}(a+1)\right)-\frac{\pi \mathrm{\Gamma }(a+1)}{4.2^{\frac{a+1}{2}}}sin\left(\frac{\pi }{4}(a+1)\right)-\frac{log2}{2}.\frac{\mathrm{\Gamma }(a+1)}{2^{\frac{a+1}{2}}}cos\left(\frac{\pi }{4}(a+1)\right)$$$$I^{\prime }\left(5\right)=\frac{-\pi .5!}{4.2^3}sin\frac{3\pi }{2}=\frac{30\pi }{8}=\frac{15\pi }{4}$$

Answered by mathmax by abdo last updated on 26/Jan/21

$$\mathrm{F}={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^5\ln \left(\mathrm{x}\right)\mathrm{cosx}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\mathrm{let}\mathrm{F}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{a}}\mathrm{cosx}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\Rightarrow$$$$\mathrm{F}\left(\mathrm{a}\right)=\mathrm{Re}\left({\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{a}}{\mathrm{e}}^{\mathrm{ix}-\mathrm{x}}\mathrm{dx}\right)\mathrm{and}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{a}}{\mathrm{e}}^{(-1+\mathrm{i})\mathrm{x}}\mathrm{dx}$$$${=}_{(1-\mathrm{i})\mathrm{x}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{t}}{1-\mathrm{i}}\right)}^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dt}}{1-\mathrm{i}}=\frac{1}{{(1-\mathrm{i})}^{\mathrm{a}+1}}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\mathrm{\Gamma }(\mathrm{a}+1)\times \frac{1}{{\left(\sqrt{2}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}\right)}^{\mathrm{a}+1}}=\frac{\mathrm{\Gamma }(\mathrm{a}+1)}{2^{\frac{\mathrm{a}+1}{2}}}{\mathrm{e}}^{\mathrm{i}\left(\frac{\mathrm{a}+1}{4}\right)\pi }$$$$\Rightarrow \mathrm{F}\left(\mathrm{a}\right)=\frac{\mathrm{\Gamma }(\mathrm{a}+1)}{2^{\frac{\mathrm{a}+1}{2}}}\cos \left(\frac{\pi }{4}(\mathrm{a}+1)\right)(\mathrm{a}>-1)$$$$\mathrm{F}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{\mathrm{alnx}}\mathrm{cosx}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\Rightarrow {\mathrm{F}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\mathrm{lnx}.{\mathrm{x}}^{\mathrm{a}}\mathrm{cosx}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\Rightarrow$$$$\Rightarrow \mathrm{F}={\mathrm{F}}^{{}^{\prime }}\left(5\right)\mathrm{we}\mathrm{have}$$$$\mathrm{F}\left(\mathrm{a}\right)=2^{-\frac{\mathrm{a}+1}{2}}\mathrm{\Gamma }(\mathrm{a}+1)\cos (\frac{\pi }{4}\mathrm{a}+\frac{\pi }{4})\Rightarrow$$$${\mathrm{F}}^{{}^{\prime }}\left(\mathrm{a}\right)={\left(2^{-\frac{\mathrm{a}+1}{2}}\mathrm{\Gamma }(\mathrm{a}+1)\right)}^{{}^{\prime }}\cos (\frac{\pi \mathrm{a}}{4}+\frac{\pi }{4})-\frac{\pi }{4}{2}^{-\frac{\mathrm{a}+1}{2}}\mathrm{\Gamma }(\mathrm{a}+1)\sin (\frac{\pi \mathrm{a}}{4}+\frac{\pi }{4})$$$$2^{-\frac{\mathrm{a}+1}{2}}\mathrm{\Gamma }(\mathrm{a}+1)={\mathrm{e}}^{-\frac{\mathrm{a}+1}{2}\ln \left(2\right)}\mathrm{\Gamma }(\mathrm{a}+1)\Rightarrow$$$$\frac{\mathrm{d}}{\mathrm{da}}(...)=-\frac{\ln 2}{2}.2^{-\frac{\mathrm{a}+1}{2}}.\mathrm{\Gamma }(\mathrm{a}+1)+{\mathrm{e}}^{-\frac{\mathrm{a}+1}{2}}{\mathrm{\Gamma }}^{{}^{\prime }}(\mathrm{a}+1)=....$$$$\mathrm{be}\mathrm{continued}...$$

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