# Integration

• Nov 15th 2013, 05:10 AM
Darrylcwc
Integration
• Nov 15th 2013, 06:00 AM
SlipEternal
Re: Integration
\begin{align*}\log(e^{2t}+5) & = \log \left[ 5 \left( \dfrac{ e^{2t} }{5} + 1 \right) \right] \\ & = \log(5) + \log\left( \dfrac{ e^{2t} }{5} + 1 \right)\end{align*}

So, $\int_2^x \log(e^{2t}+5)dt = (x-2)\log(5) + \int_2^x \left( \sum_{n\ge 1} (-1)^{n+1}\dfrac{e^{2nt}}{n5^n} \right)dt$

Note: I replaced $\log\left(\dfrac{e^{2t}}{5}+1\right)$ with its power series. The radius of convergence of the power series is $\left| \dfrac{e^{2t} }{5} \right| < 1$, but over the complex numbers, it can be analytically extended uniquely to the original function. Similarly, once you integrate, the radius of convergence of the series you obtain will also only be finite, but can also be analytically extended (uniquely) over the complex numbers to an expression containing the dilogarithm function (the polylogarithm function where $n=2$).
• Nov 16th 2013, 02:37 AM
Darrylcwc
Re: Integration
Could you break down how you arrived at the 3rd line of reasoning from the second?
• Nov 16th 2013, 05:30 AM
SlipEternal
Re: Integration
Quote:

Originally Posted by Darrylcwc
Could you break down how you arrived at the 3rd line of reasoning from the second?

The paragraph directly below the third line explains in a lot of detail precisely how I got from the second line to the third line. Which part is confusing you?
• Nov 16th 2013, 06:29 AM
Darrylcwc
Re: Integration
I think I got it.