Question Number 83078 by mhmd last updated on 27/Feb/20
$$\mathrm{If}\:\:\:{I}_{\mathrm{1}} =\underset{{e}} {\overset{{e}^{\mathrm{2}} } {\int}}\:\frac{{dx}}{\mathrm{log}\:{x}}\:\:\mathrm{and}\:\:{I}_{\mathrm{2}} =\:\underset{\:\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{{e}^{{x}} }{{x}}\:{dx},\:\mathrm{then} \\ $$
Commented by mathmax by abdo last updated on 27/Feb/20
![I_1 =_(lnx=t) ∫_1 ^2 ((e^t dt)/t) ⇒ I_1 =I_2 at form of serie I_1 =∫_1 ^2 (1/t)(Σ_(n=0) ^∞ (t^n /(n!)))dt =∫_1 ^2 (dt/t) +∫_1 ^2 Σ_(n=1) ^∞ (t^(n−1) /(n!))dt =ln(2)+Σ_(n=1) ^∞ (1/(n!)) ∫_1 ^2 t^(n−1) dt =ln(2)+Σ_(n=1) ^∞ (1/(n!))×[(1/n)t^n ]_1 ^2 =ln(2)+Σ_(n=1) ^∞ (1/(n×n!))(2^n −1)](https://www.tinkutara.com/question/Q83081.png)
$${I}_{\mathrm{1}} =_{{lnx}={t}} \:\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{e}^{{t}} \:{dt}}{{t}}\:\Rightarrow\:{I}_{\mathrm{1}} ={I}_{\mathrm{2}} \\ $$$${at}\:{form}\:{of}\:{serie}\:{I}_{\mathrm{1}} =\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{1}}{{t}}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{t}^{{n}} }{{n}!}\right){dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dt}}{{t}}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}−\mathrm{1}} }{{n}!}{dt}\:={ln}\left(\mathrm{2}\right)+\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}!}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:{t}^{{n}−\mathrm{1}} \:{dt} \\ $$$$={ln}\left(\mathrm{2}\right)+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}×\left[\frac{\mathrm{1}}{{n}}{t}^{{n}} \right]_{\mathrm{1}} ^{\mathrm{2}} \:={ln}\left(\mathrm{2}\right)+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}×{n}!}\left(\mathrm{2}^{{n}} −\mathrm{1}\right) \\ $$