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Benedict Irwin edited untitled.tex
over 8 years ago
Commit id: e0ad1ec7da93097ba8129ec9de300d39cdc9db1d
deletions | additions
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index 06de4bf..a9eb607 100644
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s_2=-\int_{-\infty}^{\infty}\frac{e^{-x^2}(16x^4-16x^2+4)}{8\sqrt{\pi}}\left(\sum_{n=1}^\infty \frac{1}{n}\left(\frac{e^{-x^2}(16x^4-16x^2+4)}{8\sqrt{\pi}} -1 \right)^n \right)\;dx
\end{equation}
Note down each term in the converging sequence
\begin{align}
1-\frac{41}{64\sqrt{2}\pi^{1/2}}\\
\frac{1}{2}+\frac{11}{54\sqrt{3}\pi}-\frac{41}{64\sqrt{2}\pi^{1/2}}\\
\frac{1}{3}-\frac{272753}{3145728\sqrt{4}\pi^{3/2}}+2\frac{11}{54\sqrt{3}\pi}-\frac{41}{64\sqrt{2}\pi^{1/2}}\\
\frac{1}{4}+\frac{65209}{1562500\sqrt{5}\pi^2}-3\frac{272753}{3145728\sqrt{4}\pi^{3/2}}+3\frac{11}{54\sqrt{3}\pi}-\frac{41}{64\sqrt{2}\pi^{1/2}}\\
\end{align}