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sdvanced-cslculus-if-x-R-and-x-0-x-e-t-1-t-ln-x-t-dt-then-prove-that-0-e-x-x-dx-2-




Question Number 136381 by mnjuly1970 last updated on 21/Mar/21
             ......sdvanced   cslculus......   if  x∈R^+  and::                 𝛗(x)=∫_0 ^( x) ((e^t βˆ’1)/t)ln((x/t))dt    then prove  that ::               Ξ¨=∫_0 ^( ∞) e^(βˆ’x) 𝛗(x)dx=ΞΆ(2)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:……{sdvanced}\:\:\:{cslculus}…… \\ $$$$\:{if}\:\:{x}\in\mathbb{R}^{+} \:{and}::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\left({x}\right)=\int_{\mathrm{0}} ^{\:{x}} \frac{{e}^{{t}} βˆ’\mathrm{1}}{{t}}{ln}\left(\frac{{x}}{{t}}\right){dt} \\ $$$$\:\:{then}\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Psi=\int_{\mathrm{0}} ^{\:\infty} {e}^{βˆ’{x}} \boldsymbol{\phi}\left({x}\right){dx}=\zeta\left(\mathrm{2}\right) \\ $$
Answered by Ñï= last updated on 21/Mar/21
Ξ¦(x)=∫_0 ^x ((e^t βˆ’1)/t)ln((x/t))dt  =∫_0 ^1 ((1βˆ’e^(xt) )/t)lntdt  =∫_0 ^1 ((1βˆ’e^(x(1βˆ’t)) )/(1βˆ’t))ln(1βˆ’t)dt  Ξ¨=∫_0 ^∞ e^(βˆ’x) ∫_0 ^1 ((1βˆ’e^(x(1βˆ’t)) )/(1βˆ’t))ln(1βˆ’t)dtdx  =∫_0 ^∞ ∫_0 ^1 ((e^(βˆ’x) βˆ’e^(βˆ’xt) )/(1βˆ’t))ln(1βˆ’t)dtdx  =∫_0 ^1 ((1βˆ’(1/t))/(1βˆ’t))ln(1βˆ’t)dt  =βˆ’βˆ«_0 ^1 ((ln(1βˆ’t))/t)dt  =Li_2 (1)
$$\Phi\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{e}^{{t}} βˆ’\mathrm{1}}{{t}}{ln}\left(\frac{{x}}{{t}}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{e}^{{xt}} }{{t}}{lntdt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{e}^{{x}\left(\mathrm{1}βˆ’{t}\right)} }{\mathrm{1}βˆ’{t}}{ln}\left(\mathrm{1}βˆ’{t}\right){dt} \\ $$$$\Psi=\int_{\mathrm{0}} ^{\infty} {e}^{βˆ’{x}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{e}^{{x}\left(\mathrm{1}βˆ’{t}\right)} }{\mathrm{1}βˆ’{t}}{ln}\left(\mathrm{1}βˆ’{t}\right){dtdx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{βˆ’{x}} βˆ’{e}^{βˆ’{xt}} }{\mathrm{1}βˆ’{t}}{ln}\left(\mathrm{1}βˆ’{t}\right){dtdx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{{t}}}{\mathrm{1}βˆ’{t}}{ln}\left(\mathrm{1}βˆ’{t}\right){dt} \\ $$$$=βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}βˆ’{t}\right)}{{t}}{dt} \\ $$$$={Li}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$
Commented by mnjuly1970 last updated on 21/Mar/21
thank you sir...grateful..
$${thank}\:{you}\:{sir}…{grateful}.. \\ $$

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