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Question Number 99234 by abdomathmax last updated on 19/Jun/20
calculate ∫_0 ^∞  e^(−x) lnx dx
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}} \mathrm{lnx}\:\mathrm{dx} \\ $$
Answered by maths mind last updated on 19/Jun/20
Γ(x)=∫_0 ^∞ e^(−t) t^(x−1) dt  Γ′(x)=∫_0 ^(+∞) e^(−t) ln(t)t^(x−1) dt  Γ′(1)=∫_0 ^(+∞) e^(−t) ln(t)dt=−γ
$$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {t}^{{x}−\mathrm{1}} {dt} \\ $$$$\Gamma'\left({x}\right)=\int_{\mathrm{0}} ^{+\infty} {e}^{−{t}} {ln}\left({t}\right){t}^{{x}−\mathrm{1}} {dt} \\ $$$$\Gamma'\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{+\infty} {e}^{−{t}} {ln}\left({t}\right){dt}=−\gamma \\ $$

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