Question

find the difinite integral from zero to infinity x^3/e^x-1dx

Answer 1

$$\begin{array}{l} \frac{x^{3}}{e^{x} - 1} = \frac{e^{-x} x^{3}}{1 - e^{-x}} = e^{-x} x^{3} \sum_{k=0}^{\infty} e^{-kx} = \sum_{k=1}^{\infty} x^{3} e^{-kx} \\ \int_{0}^{\infty} x^{3} e^{-kx} dx = -\frac{1}{k} \int_{0}^{\infty} x^{3} d e^{-kx} = -\frac{1}{k} \left(-3 \int_{0}^{\infty} x^{2} e^{-kx} dx\right) = \\ = -\frac{3}{k^{2}} \int_{0}^{\infty} x^{2} d e^{-kx} = -\frac{3}{k^{2}} \left(-2 \int_{0}^{\infty} x e^{-kx} dx\right) = \\ = -\frac{6}{k^{3}} \int_{0}^{\infty} x d e^{-kx} = \frac{6}{k^{3}} \int_{0}^{\infty} e^{-kx} dx = \frac{6}{k^{4}} \\ \int_{0}^{\infty} \frac{x^{3} dx}{e^{x} - 1} = \sum_{k=1}^{\infty} \int_{0}^{\infty} x^{3} e^{-kx} dx = 6 \sum_{k=1}^{\infty} \frac{1}{k^{4}} = 6 \zeta(4) = 6 \frac{\pi^{4}}{90} = \frac{\pi^{4}}{15} \end{array}$$

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