This cropped up on Reddit:

$I = \int_0^1 \br { \Pi_{r=1}^{10}(x+r)}\br{\sum_{r=1}^{10}\frac{1}{x+r}} \dx$

Show that $I=(a)(b!)$, where $a$ and $b$ are positive integers to be found.

Hm. Tricky. I have some ideas, but I’m going to tackle it a bit more brutally at first.

What would happen if the top limit was something other than 10, like, say, 1?

$I_1 = \int_0^1 (x+1) \frac{1}{x+1} \dx = 1$. So that’s not much to write home about.

$I_2 = \int_0^1 (x+1)(x+2) \br{\frac{1}{x+1} + \frac{1}{x+2} } \dx$

That’s more interesting. It’s $\int_0^1 \br{2x+3}\dx$, or $[x^2+3x]_0^1$, which is 4.

Is there going to be some sort of pattern when the sum of fractions gets added? I fear $I_3$ is going to be messy, but I can see what the pattern looks like:

$I_3 = \int_0^1 (x+1)(x+2) + (x+2)(x+3) + (x+3)(x+1) \dx$

$\dots = \int_0^1 3x^2 + 12x + 11 \dx$, or $[x^3 + 6x^2 + 11x]_0^1$, which is 18.

Is there a pattern forming? I can see $I_n = (n)(n!)$ as a possibility… but I don’t see a way to prove it straight away.

What if I write the original integral as $\int_0^1 P S \dx$, where $P$ is the product and $S$ the sum?
Then $\diff{P}{x} = S \br{\frac{1}{x+1} + \frac{1}{x+2} + \dots + \frac{1}{x+10}} = SP$.
Oo! That’s accidentally done the integration for us! $\int SP dx = P$ ((Plus a constant.))
So, $I = [\Pi_{r=1}^{10}(x+r)]_0^1$, which is $11! - 10!$, or $(10)(10!)$ as we suspected.