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Benedict Irwin edited untitled.tex
over 9 years ago
Commit id: ae122bead1b3e4d63e02c86b0de32a7d00fec20c
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\end{equation}
Our postulated result. Amazingly enough it is correct! (when verified elsewhere).
We can try another result, potentially as \begin{equation}
\int_0^\infty \frac{x^3}{e^x-1} dx = \frac{\pi^4}{15}
\end{equation}
We could differentiate the integrand and times by x, then flip the sign and gain the result \begin{equation}
\int_0^\infty \frac{e^x(x-3)+3)x^3}{e^x-1)^2 dx = \frac{\pi^4}{15}
\end{equation}