http://www3.wolframalpha.com/Calcula...6&w=117.&h=35.

Can someone help me with this?

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- Nov 15th 2013, 05:10 AMDarrylcwcIntegration
http://www3.wolframalpha.com/Calcula...6&w=117.&h=35.

Can someone help me with this? - Nov 15th 2013, 06:00 AMSlipEternalRe: Integration
$\displaystyle \begin{align*}\log(e^{2t}+5) & = \log \left[ 5 \left( \dfrac{ e^{2t} }{5} + 1 \right) \right] \\ & = \log(5) + \log\left( \dfrac{ e^{2t} }{5} + 1 \right)\end{align*}$

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

Note: I replaced $\displaystyle \log\left(\dfrac{e^{2t}}{5}+1\right)$ with its power series. The radius of convergence of the power series is $\displaystyle \left| \dfrac{e^{2t} }{5} \right| < 1$, 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 $\displaystyle n=2$). - Nov 16th 2013, 02:37 AMDarrylcwcRe: Integration
Could you break down how you arrived at the 3rd line of reasoning from the second?

- Nov 16th 2013, 05:30 AMSlipEternalRe: Integration
- Nov 16th 2013, 06:29 AMDarrylcwcRe: Integration
I think I got it.