how do I evaluate the integral (e^x)/x as an infinite series? thank you?

Answer 1

e^x = 1 + x + x^2/(2!) + x^3/(3!) + ... + x^n/(n!) + ...
example: 5! = 1*2*3*4*5.
(e^x)/x = 1/x + 1 + x/(2!) + x^2/(3!) + x^3/(4!) + .. + x^(n-1)/(n!) + ... That's it !!

Answer 2

e^x = 1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120 + ...
= SUM x^n/n!

Divide by x to find

e^x/x = 1/x + 1 + x/2 + ... = 1/x + SUM x^n/(n+1)!

The integral of 1/x is ln x; the integral of x^n is x^(n+1) / (n+1). So we find

INT e^x/x = ln x + SUM x^(n+1) / (n+1)(n+1)! + C
with n >= 2; or

INT e^x/x = ln x + SUM x^n / n . n! + C
with n>=1.

= ln x + x + x^2/4 + x^3/18 + x^4/96 + ... + C

Answer 3

the integral(e^x)/x as an infinite series would be the summation from n=1 to infinite of (e^n)/n.

When you say evaluate are you asking if the infinite series converges or diverges.