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Question Number 172352 by Mikenice last updated on 25/Jun/22

Answered by Mathspace last updated on 26/Jun/22

$${\int }_0^{\mathrm{\infty }}{x}^6{e}^{-x}cosxdx=Re\left({\int }_0^{\mathrm{\infty }}{x}^6{e}^{-x+ix}dx\right)$$$$and{\int }_0^{\mathrm{\infty }}{x}^6{e}^{-(1-i)x}dx$$$${=}_{(1-i)x=t}{\int }_0^{\mathrm{\infty }}\frac{t^6}{{(1-i)}^6}{e}^{-t}\frac{dt}{1-i}$$$$=\frac{1}{{(1-i)}^7}{\int }_0^{\mathrm{\infty }}{t}^{7-1}{e}^{-t}dt$$$$={(1-i)}^{-7}\mathrm{\Gamma }\left(7\right)$$$$={\left(\sqrt{2}{e}^{-\frac{i\pi }{4}}\right)}^{-7}6!$$$$=\frac{6!}{{\left(\sqrt{2}\right)}^7}{e}^{i\frac{7\pi }{4}}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{x}^6{e}^{-x}cosxdx=\frac{6!}{{\left(\sqrt{2}\right)}^7}cos\left(\frac{7\pi }{4}\right)$$

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