Question Number 141417 by mnjuly1970 last updated on 18/May/21
$$\:\:\:\:\:\:\:\:\:\:……..\:{advanced}\:…\:…\:…\:{calculus}……. \\ $$$$\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\mathscr{F}:=\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \frac{{e}^{{x}} +{e}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}}{{x}}\:{dx}\overset{??} {=}\gamma \\ $$
Answered by mnjuly1970 last updated on 27/May/21
![:=∫_(−1 ) ^( 0) (e^(1/x) /x)dx=^((1/x)=−y) ∫_1 ^( ∞) −ye^(−y) (dy/y^2 ) := −∫_1 ^( ∞) (e^(−y) /y)dy=−∫_1 ^( ∞) e^(−y) ln(y)−[ln(y)e^(−y) ]_1 ^∞ :=−∫_1 ^( ∞) e^(−y) ln(y)dy⇒−∫_0 ^( ∞) e^(−y) ln(y)dy+∫_0 ^( 1) e^(−y) ln(y)dy ∫_(−1) ^( 0) (e^(1/x) /x)dx:= γ+∫_0 ^( 1) e^(−x) ln(x)dx (★) := γ +∫_0 ^( 1) ln(x)d(1−e^(−x) ) := γ+[ln(x)(1−e^(−x) )]_0 ^( 1) −∫_0 ^( 1) (((1−e^(−x) )/x))dx := γ −∫_0 ^( 1) ((1−e^(−x) )/x)dx=γ+∫_0 ^( −1) ((1−e^x )/(−x))dx :=γ +∫_(−1) ^( 0) ((1−e^x )/x) dx (★) :: ∫_(−1) ^( 0) ((e^(1/x) +e^x −1)/x) =γ .....✓ ........ F := ∫_(−1) ^( 0) ((e^x +e^(1/x) −1)/x) dx = γ ......](https://www.tinkutara.com/question/Q142213.png)
$$\:\:\:\::=\int_{−\mathrm{1}\:} ^{\:\mathrm{0}} \frac{{e}^{\frac{\mathrm{1}}{{x}}} }{{x}}{dx}\overset{\frac{\mathrm{1}}{{x}}=−{y}} {=}\int_{\mathrm{1}} ^{\:\infty} −{ye}^{−{y}} \frac{{dy}}{{y}^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\::=\:−\int_{\mathrm{1}} ^{\:\infty} \frac{{e}^{−{y}} }{{y}}{dy}=−\int_{\mathrm{1}} ^{\:\infty} {e}^{−{y}} {ln}\left({y}\right)−\left[{ln}\left({y}\right){e}^{−{y}} \right]_{\mathrm{1}} ^{\infty} \\ $$$$\:\:\:\:\::=−\int_{\mathrm{1}} ^{\:\infty} {e}^{−{y}} {ln}\left({y}\right){dy}\Rightarrow−\int_{\mathrm{0}} ^{\:\infty} {e}^{−{y}} {ln}\left({y}\right){dy}+\int_{\mathrm{0}} ^{\:\mathrm{1}} {e}^{−{y}} {ln}\left({y}\right){dy} \\ $$$$\:\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \frac{{e}^{\frac{\mathrm{1}}{{x}}} }{{x}}{dx}:=\:\gamma+\int_{\mathrm{0}} ^{\:\mathrm{1}} {e}^{−{x}} {ln}\left({x}\right){dx}\:\left(\bigstar\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\::=\:\gamma\:+\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({x}\right){d}\left(\mathrm{1}−{e}^{−{x}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\::=\:\gamma+\left[{ln}\left({x}\right)\left(\mathrm{1}−{e}^{−{x}} \right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}−{e}^{−{x}} }{{x}}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\::=\:\gamma\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{e}^{−{x}} }{{x}}{dx}=\gamma+\int_{\mathrm{0}} ^{\:−\mathrm{1}} \frac{\mathrm{1}−{e}^{{x}} }{−{x}}{dx}\: \\ $$$$\:\:\:\:\:\:\:\:\:\::=\gamma\:+\int_{−\mathrm{1}} ^{\:\mathrm{0}} \frac{\mathrm{1}−{e}^{{x}} }{{x}}\:{dx} \\ $$$$\:\:\:\:\:\:\left(\bigstar\right)\:::\:\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \frac{{e}^{\frac{\mathrm{1}}{{x}}} +{e}^{{x}} −\mathrm{1}}{{x}}\:=\gamma\:…..\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……..\:\:\:\mathscr{F}\::=\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \frac{{e}^{{x}} +{e}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}}{{x}}\:{dx}\:=\:\gamma\:…… \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by mindispower last updated on 19/May/21
![x=(1/t)⇒F=∫_(−1) ^0 ((e^x −1)/x)dx+∫_(−1) ^0 (e^(1/x) /x)dx=A+B A=[e^x −1]ln(−x)]_(−1) ^0 −∫_(−1) ^0 e^x ln(−x) =−∫_(−1) ^0 e^x ln(−x)dx=∫_1 ^0 e^(−t) ln(t)dt B=[ln(−x)e^(1/x) ]_(−1) ^0 −∫_(−1) ^0 (e^(1/x) /x^2 )ln(−x)dx =∫_(−1) ^0 −(e^(1/x) /x^2 )ln(−x),t=−(1/x)⇒ B=−∫_1 ^∞ e^(−t) ln(t) F=A+B=−∫_0 ^∞ ln(t)e^(−t) dt=∂_x (−∫_0 ^∞ e^(−t) t^(x−1) dt)∣_(x=1) =∂_x .−Γ(x+1)∣_(x=0) =−Γ′(1)=−Γ(1)Ψ(1)=−1.−γ=γ ⇒∫_(−1) ^0 ((e^x +e^(1/x) −1)/x)dx=γ](https://www.tinkutara.com/question/Q141514.png)
$${x}=\frac{\mathrm{1}}{{t}}\Rightarrow\mathscr{F}=\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{{e}^{{x}} −\mathrm{1}}{{x}}{dx}+\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{{e}^{\frac{\mathrm{1}}{{x}}} }{{x}}{dx}={A}+{B} \\ $$$$\left.{A}=\left[{e}^{{x}} −\mathrm{1}\right]{ln}\left(−{x}\right)\right]_{−\mathrm{1}} ^{\mathrm{0}} −\int_{−\mathrm{1}} ^{\mathrm{0}} {e}^{{x}} {ln}\left(−{x}\right) \\ $$$$=−\int_{−\mathrm{1}} ^{\mathrm{0}} \boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{ln}}\left(−\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\int_{\mathrm{1}} ^{\mathrm{0}} {e}^{−{t}} {ln}\left({t}\right){dt} \\ $$$$\boldsymbol{{B}}=\left[{ln}\left(−{x}\right){e}^{\frac{\mathrm{1}}{{x}}} \right]_{−\mathrm{1}} ^{\mathrm{0}} −\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{{e}^{\frac{\mathrm{1}}{{x}}} }{{x}^{\mathrm{2}} }{ln}\left(−{x}\right){dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} −\frac{\boldsymbol{{e}}^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} }{\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{ln}}\left(−\boldsymbol{{x}}\right),{t}=−\frac{\mathrm{1}}{{x}}\Rightarrow \\ $$$${B}=−\int_{\mathrm{1}} ^{\infty} {e}^{−{t}} {ln}\left({t}\right) \\ $$$$\mathscr{F}={A}+{B}=−\int_{\mathrm{0}} ^{\infty} {ln}\left({t}\right){e}^{−{t}} {dt}=\partial_{{x}} \left(−\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {t}^{{x}−\mathrm{1}} {dt}\right)\mid_{{x}=\mathrm{1}} \\ $$$$=\partial_{{x}} .−\Gamma\left({x}+\mathrm{1}\right)\mid_{{x}=\mathrm{0}} =−\Gamma'\left(\mathrm{1}\right)=−\Gamma\left(\mathrm{1}\right)\Psi\left(\mathrm{1}\right)=−\mathrm{1}.−\gamma=\gamma \\ $$$$\Rightarrow\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{{e}^{{x}} +{e}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}}{{x}}{dx}=\gamma \\ $$$$ \\ $$$$ \\ $$