Summation form of improper integrals

Question

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.

Answer 1

So, starting at $$ \int_0^{\infty} e^{-nx} x^{s-1} \, \mathrm{d}x = \frac{\Pi(s-1)}{n^s} $$ we sum up over $n \in \mathbb{N}$. That is, we investigate $$ \mathcal{L} := \sum_{n=1}^\infty \int_0^{\infty} e^{-nx} x^{s-1} \, \mathrm{d}x = \sum_{n=1}^\infty \frac{\Pi(s-1)}{n^s} =: \mathcal{R} \tag{1} $$ The right-hand side is easy to establish since $\Pi(s-1)$ is independent of $n$, and (if $s>1$) the summation is known to be convergent (e.g. via the integral test): $$\mathcal{R} = \sum_{n=1}^\infty \frac{\Pi(s-1)}{n^s} = \Pi(s-1)\sum_{n=1}^\infty \frac{1}{n^s} = \Pi(s-1) \zeta(s) $$ For the left-hand side of $(1)$, suppose the interchange of sum and integral is valid. Then $$ \mathcal{L} = \sum_{n=1}^\infty \int_0^{\infty} e^{-nx} x^{s-1} \, \mathrm{d}x \stackrel{\ast}{=} \int_0^{\infty} \sum_{n=1}^\infty e^{-nx} x^{s-1} \, \mathrm{d}x = \int_0^{\infty} x^{s-1} \sum_{n=1}^\infty (e^{-x})^n\, \mathrm{d}x $$ Here, the second equality is the interchange mentioned, and the third is factoring out the $x^{s-1}$ term. Note that the summation here converges, since $e^{-x} \in (0,1)$ for $x \in (0,\infty)$, and is a geometric series in the ratio $e^{-x}$, so $$ \mathcal{L} = \int_0^\infty x^{s-1} \cdot \frac{e^{-x}}{1-e^{-x}}\, \mathrm{d}x $$ Multiplying the top and bottom by $e^x$ gives $$ \frac{e^{-x}}{1-e^{-x}} = \frac{e^{-x}}{1-e^{-x}} \cdot \frac{e^x}{e^x} = \frac{1}{e^x - 1} $$ so $$ \mathcal{L} = \int_0^\infty \frac{x^{s-1}}{e^x-1} \, \mathrm{d}x $$ as desired.

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Above, an equality was starred. This was because it featured the interchange of integral and summation, which need not always hold. A standard result is that $$ \sum_{n=N}^\infty \int_a^\infty f_n(x) \, \mathrm{d}x = \int_a^\infty \sum_{n=N}^\infty f_n(x) \, \mathrm{d}x $$ if:

• $f_n$ is locally integrable on $[a,\infty)$ for each $n$, i.e. integrable over any compact subset
• the sum converges uniformly (for which tests like the Weierstrass $M$-test often work well)
• there is an integrable function $\varphi$ such that $\int_a^\infty \varphi(x) \, \mathrm{d}x < \infty$ and $|f_n| \le \varphi$.

I'll leave the details justifying the interchange for you.

Answer 2

Hope this is what you are looking for:

$$I = \displaystyle\int_0^{\infty} \frac{x^{s-1}}{e^x - 1} dx = \displaystyle\int_0^{\infty} \frac{e^{-x}x^{s-1}}{1 - e^{-x}} dx = \displaystyle\int_0^{\infty} \sum_{j=0}^{\infty}e^{-xj}e^{-x}x^{s-1}dx$$

$$ I = \int_0^{\infty} \sum_{j=0}^{\infty}e^{-x(j+1)}x^{s-1}dx = \underbrace{\sum_{j=0}^{\infty}\int_0^{\infty} e^{-x(j+1)}x^{s-1}dx = \sum_{j=0}^{\infty}\frac{\Pi(s-1)}{(j+1)^s}}_{\textrm{page 9 expression}}$$

$$ I = \Pi(s-1)\sum_{j=0}^{\infty}\frac{1}{(j+1)^s} = \Pi(s-1)\zeta(s)$$