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Question Number 85721 by M±th+et£s last updated on 24/Mar/20

show that ∫_0 ^∞ ((e^(−x) ln(x))/(√x))dx=−(√π)(γ+ln(4))

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} {ln}\left({x}\right)}{\sqrt{{x}}}{dx}=−\sqrt{\pi}\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$

Answered by mind is power last updated on 24/Mar/20

∫_0 ^(+∞) ((e^(−x) ln(x))/(√x))dx (√x)=u ⇒∫_0 ^(+∞) 4e^(−u^2 ) ln(u)du=4∫_0 ^(+∞) e^(−u^2 ) ln(u)du ∫_0 ^(+∞) t^(x−1) e^(−t) dt=Γ(x) t=u^2 ⇒dt=2udu 2∫_0 ^(+∞) u^(2x−1) e^(−u^2 ) =Γ(x)⇒Γ′(x)=2∫2ln(u)u^(2x−1) e^(−u^2 ) du=Γ′(x) ⇒4∫_0 ^(+∞) ln(u)e^(−u^2 ) du=Γ′((1/2))=Ψ((1/2))Γ((1/2)) Ψ((1/2))=−γ−ln(2.2)=−γ−ln(4) Γ((1/2))=(√π)⇒ ∫_0 ^(+∞) ((e^(−x) ln(x))/(√x))dx=4∫_0 ^(+∞) ln(u)e^(−u^2 ) du=(√π)(−γ−ln(4))=−(√π)(γ+ln(4))

$$\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}} {ln}\left({x}\right)}{\sqrt{{x}}}{dx} \\ $$$$\sqrt{{x}}={u} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \mathrm{4}{e}^{−{u}^{\mathrm{2}} } {ln}\left({u}\right){du}=\mathrm{4}\int_{\mathrm{0}} ^{+\infty} {e}^{−{u}^{\mathrm{2}} } {ln}\left({u}\right){du} \\ $$$$\int_{\mathrm{0}} ^{+\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}=\Gamma\left({x}\right) \\ $$$${t}={u}^{\mathrm{2}} \Rightarrow{dt}=\mathrm{2}{udu} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {u}^{\mathrm{2}{x}−\mathrm{1}} {e}^{−{u}^{\mathrm{2}} } =\Gamma\left({x}\right)\Rightarrow\Gamma'\left({x}\right)=\mathrm{2}\int\mathrm{2}{ln}\left({u}\right){u}^{\mathrm{2}{x}−\mathrm{1}} {e}^{−{u}^{\mathrm{2}} } {du}=\Gamma'\left({x}\right) \\ $$$$\Rightarrow\mathrm{4}\int_{\mathrm{0}} ^{+\infty} {ln}\left({u}\right){e}^{−{u}^{\mathrm{2}} } {du}=\Gamma'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma−{ln}\left(\mathrm{2}.\mathrm{2}\right)=−\gamma−{ln}\left(\mathrm{4}\right) \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}} {ln}\left({x}\right)}{\sqrt{{x}}}{dx}=\mathrm{4}\int_{\mathrm{0}} ^{+\infty} {ln}\left({u}\right){e}^{−{u}^{\mathrm{2}} } {du}=\sqrt{\pi}\left(−\gamma−{ln}\left(\mathrm{4}\right)\right)=−\sqrt{\pi}\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$

Commented by M±th+et£s last updated on 24/Mar/20

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$

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