Question-172352

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

Answered by Mathspace last updated on 26/Jun/22

∫_0 ^∞ x^6 e^(−x) cosx dx=Re(∫_0 ^∞ x^6 e^(−x+ix) dx) and ∫_0 ^∞ x^6 e^(−(1−i)x) dx =_((1−i)x=t) ∫_0 ^∞ (t^6 /((1−i)^6 )) e^(−t) (dt/(1−i)) =(1/((1−i)^7 ))∫_0 ^∞ t^(7−1) e^(−t) dt =(1−i)^(−7) Γ(7) =((√2)e^(−((iπ)/4)) )^(−7) 6! =((6!)/(((√2))^7 ))e^(i((7π)/4)) ⇒ ∫_0 ^∞ x^6 e^(−x) cosx dx=((6!)/(((√2))^7 ))cos(((7π)/4))

$$\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{6}} {e}^{−{x}} {cosx}\:{dx}={Re}\left(\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{6}} {e}^{−{x}+{ix}} {dx}\right)\: \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{6}} {e}^{−\left(\mathrm{1}−{i}\right){x}} {dx} \\ $$$$=_{\left(\mathrm{1}−{i}\right){x}={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{6}} }{\left(\mathrm{1}−{i}\right)^{\mathrm{6}} }\:{e}^{−{t}} \frac{{dt}}{\mathrm{1}−{i}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{i}\right)^{\mathrm{7}} }\int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{7}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$=\left(\mathrm{1}−{i}\right)^{−\mathrm{7}} \Gamma\left(\mathrm{7}\right) \\ $$$$=\left(\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{7}} \mathrm{6}! \\ $$$$=\frac{\mathrm{6}!}{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} }{e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{6}} \:{e}^{−{x}} {cosx}\:{dx}=\frac{\mathrm{6}!}{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} }{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{4}}\right) \\ $$

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Question-172352

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