Question

Evaluate\({\rm{ }}\int {\frac{{{{\rm{e}}^{\rm{x}}}}}{{\rm{x}}}} {\rm{dx }}\) as an infinite series.

Short answer

The value of \(\int {\frac{{{{\rm{e}}^{\rm{x}}}}}{{\rm{x}}}} {\rm{dx }}\)is \(C + \ln |x| + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n(n!)}}} \).

Step-by-step solution

Definition of Concept

Infinite series:The sum of an unlimited number of numbers connected in a specific fashion and listed in a specific order is known as an infinite series.

Given Terms

The\({{\rm{e}}^{\rm{x}}}\)power series is

\({e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \).

So, the \(\frac{{{{\rm{e}}^{\rm{x}}}}}{{\rm{x}}}\) power series is

\(\frac{{{e^x}}}{x} = \frac{1}{x}\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} = \sum\limits_{n = 0}^\infty {\frac{{{x^{n - 1}}}}{{n!}}} \).

Calculation

Consider the given values and simplify,

\(\begin{array}{c}\int {\frac{{{e^x}}}{x}} dx = \int {\sum\limits_{n = 0}^\infty {\frac{{{x^{n - 1}}}}{{n!}}} } dx\\ = \int {\frac{1}{x}} + \sum\limits_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{{n!}}} dx\end{array}\)

\(\begin{array}{c} = \int {\frac{1}{x}} dx + \int {\sum\limits_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{{n!}}} } dx\\ = C + \ln |x| + \sum\limits_{n = 1}^\infty {\int {\frac{{{x^{n - 1}}}}{{n!}}} } dx\\ = C + \ln |x| + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n(n!)}}} \end{array}\)

Hence, the required solution is \(C + \ln |x| + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n(n!)}}} \).