Question Number 172352 by Mikenice last updated on 25/Jun/22
Answered by Mathspace last updated on 26/Jun/22
$$\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{6}} {e}^{−{x}} {cosx}\:{dx}={Re}\left(\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{6}} {e}^{−{x}+{ix}} {dx}\right)\: \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{6}} {e}^{−\left(\mathrm{1}−{i}\right){x}} {dx} \\ $$$$=_{\left(\mathrm{1}−{i}\right){x}={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{6}} }{\left(\mathrm{1}−{i}\right)^{\mathrm{6}} }\:{e}^{−{t}} \frac{{dt}}{\mathrm{1}−{i}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{i}\right)^{\mathrm{7}} }\int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{7}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$=\left(\mathrm{1}−{i}\right)^{−\mathrm{7}} \Gamma\left(\mathrm{7}\right) \\ $$$$=\left(\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{7}} \mathrm{6}! \\ $$$$=\frac{\mathrm{6}!}{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} }{e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{6}} \:{e}^{−{x}} {cosx}\:{dx}=\frac{\mathrm{6}!}{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} }{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{4}}\right) \\ $$