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Question Number 222850 by MrGaster last updated on 09/Jul/25
∫_0 ^∞ ((x^(−x) e^(−x) )/(Γ(1−x)))dx
$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{−{x}} {e}^{−{x}} }{\Gamma\left(\mathrm{1}−{x}\right)}{dx} \\ $$
Answered by MrGaster last updated on 09/Jul/25
Γ(1−x)Γ(x)=(π/(sin(πx))) I=∫_0 ^∞ ((x^(−x) e^(−x) )/(Γ(1−x)))dx =(1/π)∫_0 ^∞ ((Γ(x)sin(πx))/(x^e e^x ))dx I=(1/π)∫_0 ^∞ ((Γ(x)sin(πx))/(x^e e^x ))dx =(1/π)∫_0 ^∞ ((sin(πx))/e^x )∫_0 ^∞ t^(x−1) e^(−tx) dtdx =(1/π)∫_0 ^∞ (1/t)∫_0 ^∞ (te^(−(t+1)) )^x sin(πx)dxdt a=te^(−(t+1)) I=(1/π)∫_0 ^∞ (1/t)∫_0 ^∞ a^x sin(πx)dxdt =(1/π)∫_0 ^∞ (π/(ln^2 (a)+π^2 )) (dt/t) =∫_0 ^∞ (1/((ln t−t+1)^2 +π^2 )) (dt/t) =∫_(−∞) ^∞ (1/((e^u −u+1)^2 +π^2 ))du u=ln t I=∫_(−∞) ^∞ (1/((e^u −u+1)^2 +π^2 ))du =∫_(−∞) ^∞ (1/((e^u −u+1+iπ)(e^u −u+1−iπ)))du =(1/(2πi))∫_(−∞) ^∞ ((1/(e^u −u+1−iπ))−(1/(e^u −u+1+iπ)))du =(1/(2πi))lim_(R→∞) I_1 −I_2 z=u+iπ and z=u−iπ Replace two variables I_1 and I_2 respectively. I_1 =∫_(−R) ^R (1/(e^u −u+1−iπ))du=∫_(R+iπ) ^(−R−iπ) (1/(−e^z −z+1))dz I_2 =∫_(−R) ^R (1/(e^u −u+1+iπ))du=∫_(−R−iπ) ^(R−iπ) (1/(−e^z −z−1))dz lim_(R→∞) ∫_(R−iπ) ^(R+iπ) (1/(−e^z −z+1))dz=0 lim_(R→∞) ∫_(−R−iπ) ^(−R+iπ) (1/(−e^z −z+1))dz=0 I=(1/(2πi))lim_(R→∞) (I_1 −I_2 ) =−(1/(2πi))∮_C (1/(−e^z −z+1))dz =Res((1/(e^z +z−1)),z=0) =(1/2)
$$\Gamma\left(\mathrm{1}−{x}\right)\Gamma\left({x}\right)=\frac{\pi}{\mathrm{sin}\left(\pi{x}\right)} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{−{x}} {e}^{−{x}} }{\Gamma\left(\mathrm{1}−{x}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\Gamma\left({x}\right)\mathrm{sin}\left(\pi{x}\right)}{{x}^{{e}} {e}^{{x}} }{dx} \\ $$$${I}=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\Gamma\left({x}\right)\mathrm{sin}\left(\pi{x}\right)}{{x}^{{e}} {e}^{{x}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\pi{x}\right)}{{e}^{{x}} }\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{tx}} {dtdx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}}\int_{\mathrm{0}} ^{\infty} \left({te}^{−\left({t}+\mathrm{1}\right)} \right)^{{x}} \mathrm{sin}\left(\pi{x}\right){dxdt} \\ $$$${a}={te}^{−\left({t}+\mathrm{1}\right)} \\ $$$${I}=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}}\int_{\mathrm{0}} ^{\infty} {a}^{{x}} \mathrm{sin}\left(\pi{x}\right){dxdt} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\pi}{\mathrm{ln}^{\mathrm{2}} \left({a}\right)+\pi^{\mathrm{2}} }\:\frac{{dt}}{{t}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{ln}\:{t}−{t}+\mathrm{1}\right)^{\mathrm{2}} +\pi^{\mathrm{2}} }\:\frac{{dt}}{{t}} \\ $$$$=\int_{−\infty} ^{\infty} \frac{\mathrm{1}}{\left({e}^{{u}} −{u}+\mathrm{1}\right)^{\mathrm{2}} +\pi^{\mathrm{2}} }{du} \\ $$$${u}=\mathrm{ln}\:{t} \\ $$$${I}=\int_{−\infty} ^{\infty} \frac{\mathrm{1}}{\left({e}^{{u}} −{u}+\mathrm{1}\right)^{\mathrm{2}} +\pi^{\mathrm{2}} }{du} \\ $$$$=\int_{−\infty} ^{\infty} \frac{\mathrm{1}}{\left({e}^{{u}} −{u}+\mathrm{1}+{i}\pi\right)\left({e}^{{u}} −{u}+\mathrm{1}−{i}\pi\right)}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{−\infty} ^{\infty} \left(\frac{\mathrm{1}}{{e}^{{u}} −{u}+\mathrm{1}−{i}\pi}−\frac{\mathrm{1}}{{e}^{{u}} −{u}+\mathrm{1}+{i}\pi}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\underset{{R}\rightarrow\infty} {\mathrm{lim}}{I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${z}={u}+{i}\pi\:\mathrm{and}\:{z}={u}−{i}\pi \\ $$$$\mathrm{Replace}\:\mathrm{two}\:\mathrm{variables}\:{I}_{\mathrm{1}} \: \\ $$$$\mathrm{and}\:{I}_{\mathrm{2}} \:\mathrm{respectively}. \\ $$$${I}_{\mathrm{1}} =\int_{−{R}} ^{{R}} \frac{\mathrm{1}}{{e}^{{u}} −{u}+\mathrm{1}−{i}\pi}{du}=\int_{{R}+{i}\pi} ^{−{R}−{i}\pi} \frac{\mathrm{1}}{−{e}^{{z}} −{z}+\mathrm{1}}{dz} \\ $$$${I}_{\mathrm{2}} =\int_{−{R}} ^{{R}} \frac{\mathrm{1}}{{e}^{{u}} −{u}+\mathrm{1}+{i}\pi}{du}=\int_{−{R}−{i}\pi} ^{{R}−{i}\pi} \frac{\mathrm{1}}{−{e}^{{z}} −{z}−\mathrm{1}}{dz} \\ $$$$\underset{{R}\rightarrow\infty} {\mathrm{lim}}\int_{{R}−{i}\pi} ^{{R}+{i}\pi} \frac{\mathrm{1}}{−{e}^{{z}} −{z}+\mathrm{1}}{dz}=\mathrm{0} \\ $$$$\underset{{R}\rightarrow\infty} {\mathrm{lim}}\int_{−{R}−{i}\pi} ^{−{R}+{i}\pi} \frac{\mathrm{1}}{−{e}^{{z}} −{z}+\mathrm{1}}{dz}=\mathrm{0} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\underset{{R}\rightarrow\infty} {\mathrm{lim}}\left({I}_{\mathrm{1}} −{I}_{\mathrm{2}} \right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\oint_{{C}} \frac{\mathrm{1}}{−{e}^{{z}} −{z}+\mathrm{1}}{dz} \\ $$$$=\mathrm{Res}\left(\frac{\mathrm{1}}{{e}^{{z}} +{z}−\mathrm{1}},{z}=\mathrm{0}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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