If-I-1-e-e-2-dx-log-x-and-I-2-1-2-e-x-x-dx-then-

Question Number 83078 by mhmd last updated on 27/Feb/20

If I_{1} = \underset{e}{\int^{e^{2}}} \frac{d x}{\log x} and I_{2} = \underset{1}{\int^{2}} \frac{e^{x}}{x} d x, then

Commented by mathmax by abdo last updated on 27/Feb/20

I_{1} =_{l n x = t} \int_{1}^{2} \frac{e^{t} d t}{t} \Rightarrow I_{1} = I_{2}

a t f o r m o f s e r i e I_{1} = \int_{1}^{2} \frac{1}{t} (\sum_{n = 0}^{\infty} \frac{t^{n}}{n!}) d t

= \int_{1}^{2} \frac{d t}{t} + \int_{1}^{2} \sum_{n = 1}^{\infty} \frac{t^{n - 1}}{n!} d t = l n (2) + \sum_{n = 1}^{\infty} \frac{1}{n!} \int_{1}^{2} t^{n - 1} d t

= l n (2) + \sum_{n = 1}^{\infty} \frac{1}{n!} \times {[\frac{1}{n} t^{n}]}_{1}^{2} = l n (2) + \sum_{n = 1}^{\infty} \frac{1}{n \times n!} (2^{n} - 1)