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[yfzhao20]

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Here $\displaystyle\int_{0}^{+\infty}\frac{x^{3}}{e^{x}-1}dx$ is just a pure number. We can calc that in the way of complex function, and the result is $\dfrac{\pi^{4}}{15}$.

<details>
<summary>How to calc $\displaystyle\int_{0}^{+\infty}\frac{x^{3}}{e^{x}-1}\mathrm{d}x$</summary>
$$
\begin{aligned}
    \int_{0}^{+\infty}\frac{x^{3}}{e^{x}-1}\mathrm{d}x&=\int_{0}^{+\infty}\frac{x^{3}e^{-x}}{1-e^{-x}}\mathrm{d}x\\
    \frac{e^{-x}}{1-e^{-x}}&=e^{-x}\sum_{n=0}^{+\infty}e^{-nx}\\
    &=\sum_{n=1}^{+\infty}e^{-nx}
\end{aligned}
$$
Obviously series $\displaystyle \sum_{n=0}^{+\infty}x^{3}e^{-nx}$ is uniformly convergent. So that:
$$
\begin{aligned}
    \int_{0}^{+\infty}\frac{x^{3}e^{-x}}{1-e^{-x}}\mathrm{d}x&=\int_{0}^{+\infty}x^{3}\sum_{n=0}^{+\infty}e^{-nx}\mathrm{d}x\\
    &=\sum_{n=0}^{+\infty}\int_{0}^{+\infty}x^{3}e^{-nx}\mathrm{d}x\\
    &=\sum_{n=0}^{+\infty}\frac{1}{n^{4}}\int_{0}^{+\infty}(nx)^{3}e^{-nx}\mathrm{d}(nx)\\
    &=\sum_{n=0}^{+\infty}\frac{1}{n^{4}}\Gamma(4)\\
    &=6\sum_{n=0}^{+\infty}\frac{1}{n^{4}}\\
\end{aligned}
$$
Here you need to calc $\displaystyle \sum_{n=0}^{+\infty}\frac{1}{n^{4}}$. You can use integration:
$$
\oint_{\small{|z|=(m+\frac{1}{2})\pi\\m\in \Z ,m\to +\infty} }^{}\frac{\mathrm{d}z}{z^{n}\tan z},n\ge 2,n\in \Z
$$
to calc $\displaystyle \sum_{n=0}^{+\infty}\frac{1}{n^{2k}},k\in \Z_{+}$. And then you can get $\displaystyle \sum_{n=0}^{+\infty}\frac{1}{n^4}=\frac{\pi^{4}}{90}$. So the result of the integration is :
$$
\int_{0}^{+\infty}\frac{x^{3}}{e^{x}-1}\mathrm{d}x =\frac{\pi^{4}}{15}
$$

---
</details>


Then we will get the final result:
$$
R(T)=\frac{2\pi^{5}k_{B}^{4}}{15h^{3}c^2}T^{4}=\sigma T^{4}
$$

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[abnerlee]

It is covered by #1939