# Definite Integral

• Jan 25th 2008, 08:36 PM
graticcio
Definite Integral
$\sum_{i=1}^n \frac{x^i}{i}$
how to change this summation into a definite integral ???

$\sum_{i=1}^n x^{i-1} = \frac{1}{1-x}$
integrate both sides, LHS will have $\sum_{i=1}^n \frac{x^i}{i}$
but what is the definite limit to take???

thanks
• Jan 25th 2008, 10:48 PM
CaptainBlack
Quote:

Originally Posted by graticcio

$\sum_{i=1}^n x^{i-1} = \frac{1}{1-x}$

This identity is wrong, to see this look at the left hand side, this is finite
for all $x$, now look at the right hand side this diverges near $x=1$.

What you have is a finite geometric series and:

$\sum_{i=1}^n x^{i-1} = \frac{1-x^n}{1-x}$

Now consider:

$\int_0^x \left[ \sum_{i=1}^n \zeta^{i-1} \right] d \zeta = \int_0^x \frac{1-\zeta^n}{1-\zeta} d \zeta$

RonL