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Question Number 72025 by mathmax by abdo last updated on 23/Oct/19
calculate ∫_0 ^∞ e^(−αx) ln(x)dx with α>0
$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\alpha{x}} {ln}\left({x}\right){dx}\:\:{with}\:\alpha>\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 24/Oct/19
changement αx=t give ∫_0 ^∞ e^(−αx) ln(x)dx=∫_0 ^∞ e^(−t) ln((t/α))(dt/α) =(1/α)∫_0 ^∞ e^(−t) {ln(t)−lnα}dt =(1/α)∫_0 ^∞ e^(−t) ln(t)−((ln(α))/α) =(1/α)(−γ)−((ln(α))/α) =−(1/α)( γ +ln(α)) ∫_0 ^∞ e^(−t) ln(t)dt=−γ and this value is proved.
$${changement}\:\alpha{x}={t}\:{give}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−\alpha{x}} {ln}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {ln}\left(\frac{{t}}{\alpha}\right)\frac{{dt}}{\alpha}\:=\frac{\mathrm{1}}{\alpha}\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} \left\{{ln}\left({t}\right)−{ln}\alpha\right\}{dt} \\ $$$$=\frac{\mathrm{1}}{\alpha}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left({t}\right)−\frac{{ln}\left(\alpha\right)}{\alpha} \\ $$$$=\frac{\mathrm{1}}{\alpha}\left(−\gamma\right)−\frac{{ln}\left(\alpha\right)}{\alpha}\:=−\frac{\mathrm{1}}{\alpha}\left(\:\gamma\:+{ln}\left(\alpha\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left({t}\right){dt}=−\gamma\:\:{and}\:{this}\:{value}\:{is}\:{proved}. \\ $$
Answered by mind is power last updated on 23/Oct/19
u=αx ⇒∫_0 ^∞ e^(−u) ln((u/α)).(du/α) =∫_0 ^(+∞) e^(−u) ln(u)(du/α)−∫_0 ^∞ e^(−u) .((ln(α))/α)du =(1/α)∫_0 ^(+∞) e^(−u) ln(u)du+((ln(α))/α)[e^(−u) ]_0 ^(+∞) =(1/α).γ−((ln(α))/α)=(1/α)(γ−ln(α))
$$\mathrm{u}=\alpha\mathrm{x} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\frac{\mathrm{u}}{\alpha}\right).\frac{\mathrm{du}}{\alpha} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\mathrm{u}\right)\frac{\mathrm{du}}{\alpha}−\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{u}} .\frac{\mathrm{ln}\left(\alpha\right)}{\alpha}\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\alpha}\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\mathrm{u}\right)\mathrm{du}+\frac{\mathrm{ln}\left(\alpha\right)}{\alpha}\left[\mathrm{e}^{−\mathrm{u}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\alpha}.\gamma−\frac{\mathrm{ln}\left(\alpha\right)}{\alpha}=\frac{\mathrm{1}}{\alpha}\left(\gamma−\mathrm{ln}\left(\alpha\right)\right) \\ $$

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