On page 9, Edwards has this expression

$$ \int_0^{\infty} e^{-nx} x^{s-1} dx = \frac{\Pi(s-1)}{n^s}$$

obtained from Euler’s factorial formula by replacing $x$ with $nx$. Can you help with the next step?

Edwards writes that

Riemann sums this over $n$ and uses

$$\sum_{n=1}^{\infty} \frac{1}{r^s} = \frac{1}{(r-1)}$$

to obtain

$$\int_0^{\infty} \frac{x^{s-1}}{e^x - 1} \,\mathrm{d}x = \Pi(s-1)\sum_{n=1}^{\infty} \frac{1}{n^s}$$

for $s>1$.

How does Riemann sum this over $n$? I asked WolframAlpha for the summation form of the integral but there was no answer.

Edwards notes that,

Convergence of the improper integral on the left and the validity of the interchange of summation and integration are not difficult to establish.

Thanks.

On page 9, Edwards has this expression

$$ \int_0^{\infty} e^{-nx} x^{s-1} dx = \frac{\Pi(s-1)}{n^s}$$

obtained from Euler’s factorial formula by replacing $x$ with $nx$. Can you help with the next step?

Edwards writes that

Riemann sums this over $n$ and uses

$$\sum_{n=1}^{\infty} \frac{1}{r^n} = \frac{1}{(r-1)}$$

to obtain

$$\int_0^{\infty} \frac{x^{s-1}}{e^x - 1} \,\mathrm{d}x = \Pi(s-1)\sum_{n=1}^{\infty} \frac{1}{n^s}$$

for $s>1$.

How does Riemann sum this over $n$? I asked WolframAlpha for the summation form of the integral but there was no answer.

Edwards notes that,

Convergence of the improper integral on the left and the validity of the interchange of summation and integration are not difficult to establish.

Thanks.

On page 9, Edwards has this expression

$\displaystyle \int_0^{\infty} e^{-nx} x^{s-1} dx = \frac{\Pi(s-1)}{n^s}$

obtained from Euler’s factorial formula by replacing $x$ with $nx$. Can you help with the next step?

Edwards writes that

Riemann sums this over $n$ and uses

$\displaystyle\sum_{n=1}^{\infty} \frac{1}{r^s} = \frac{1}{(r-1)}$

to obtain

$\displaystyle\int_0^{\infty} \frac{x^{s-1}}{e^x - 1} dx = \Pi(s-1)\sum_{n=1}^{\infty} \frac{1}{n^s}$

For $s>1$

How does Riemann sum this over $n$? I asked WolframAlpha for the summation form of the integral but there was no answer.

Edwards notes that,

Convergence of the improper integral on the left and the validity of the interchange of summation and integration are not difficult to establish.

Thanks.

On page 9, Edwards has this expression

$$ \int_0^{\infty} e^{-nx} x^{s-1} dx = \frac{\Pi(s-1)}{n^s}$$

obtained from Euler’s factorial formula by replacing $x$ with $nx$. Can you help with the next step?

Edwards writes that

Riemann sums this over $n$ and uses

$$\sum_{n=1}^{\infty} \frac{1}{r^s} = \frac{1}{(r-1)}$$

to obtain

$$\int_0^{\infty} \frac{x^{s-1}}{e^x - 1} \,\mathrm{d}x = \Pi(s-1)\sum_{n=1}^{\infty} \frac{1}{n^s}$$

for $s>1$.

How does Riemann sum this over $n$? I asked WolframAlpha for the summation form of the integral but there was no answer.

Edwards notes that,

Convergence of the improper integral on the left and the validity of the interchange of summation and integration are not difficult to establish.

Thanks.