07/02/2010, 09:35 PM
Unfortunately, I have no idea what the Taylor series is. What you linked to uses x and ln(x), while a true Taylor series would only involve x. It's be cool to see how to calculate the nth derivative of \( ^{a}x \), because then one could use the Taylor series to find the Taylor series of the integral, which would probably be simpler to calculate than my solution.
On another note, I decided to tackle \( \begin{equation}
\int^x_{0} ln(t)^{S_{n-1}} t^{i_{n-1}} dt\end{equation}
\). Here, \( S_{n-1} \) stands for \( \begin{equation} \sum^{n-1}_{k=0}i_k \end{equation} \). Integration by parts yielded \( x^{i_{n-1}+1} \sum_{k=0}^{S_{n-1}} \frac{(-1)^kS_{n-1}!ln(x)^{S_{n-1}-k}}{(S_n-k)!(i_{n-1}+1)^{k+1}} \). Thus,![[Image: bigone.jpg]](http://lh6.ggpht.com/_1md9svnyqC8/TC5MyB_fbjI/AAAAAAAAAL4/zq1H00QZ24M/bigone.jpg)
which I'm not going to bother TEXing.
Unfortunately, all those summations make calculating even \( Ti_{3}(1) \) extraordinarily difficult. Can anyone calculate this series with Maple or Mathematica? (I know that the integral approximation is ~0.573... but I want to check my series with that)
On another note, I decided to tackle \( \begin{equation}
\int^x_{0} ln(t)^{S_{n-1}} t^{i_{n-1}} dt\end{equation}
\). Here, \( S_{n-1} \) stands for \( \begin{equation} \sum^{n-1}_{k=0}i_k \end{equation} \). Integration by parts yielded \( x^{i_{n-1}+1} \sum_{k=0}^{S_{n-1}} \frac{(-1)^kS_{n-1}!ln(x)^{S_{n-1}-k}}{(S_n-k)!(i_{n-1}+1)^{k+1}} \). Thus,
which I'm not going to bother TEXing.
Unfortunately, all those summations make calculating even \( Ti_{3}(1) \) extraordinarily difficult. Can anyone calculate this series with Maple or Mathematica? (I know that the integral approximation is ~0.573... but I want to check my series with that)
