What is the antiderivative of #(e^x)/x#?
1 Answer
# int \ e^x/x \ dx = lnAx + x + x^2/(2*2!) + x^3/(3*3!) + ... x^n/(n*n!) + ... #
or in sigma notation
# int \ e^x/x \ dx = lnAx + sum_(n=1)^oo x^n/(n*n!) #
Explanation:
Let:
# I = int \ e^x/x \ dx #
This does not have an elementary solution. Definite integrals involving this integrand are calculated using tables of the Exponential integral
The best you can get is a power series which we derive from the power series of
# e^x = 1 + x + x^2/(2!) + x^3/(3!) + ... x^n/(n!) + ... + ... #
So the integral becomes:
# I = int \ 1/x{1 + x + x^2/(2!) + x^3/(3!) + ... x^n/(n!) + ... + ... } \ dx #
# \ \ = int \ 1/x + 1 + x/(2!) + x^2/(3!) + ... x^(n-1)/(n!) + ... + ... \ dx #
# \ \ = lnx + x + x^2/(2*2!) + x^3/(3*3!) + ... x^n/(n*n!) + ... + C #
# \ \ = lnAx + x + x^2/(2*2!) + x^3/(3*3!) + ... x^n/(n*n!) + ... #
# \ \ = lnAx + sum_(n=1)^oo x^n/(n*n!) #