Integral Sequence

Checking $f_{1}$ , $f_{2}$ , etc. we get to know that $f_{n} (x)$ can be expressed as $\displaystyle \frac{a_{n}}{x^{n-1}}$

$f_{n+1} x = \displaystyle \frac{1}{a_{n}} \int_{0}^{\infty} x^{n-1} e^{-xt} dt$

Using integration by parts using $e^{-xt}$ as first function, and $x^{n-1}$ as second function, we get:

$\displaystyle f_{n+1} (x) = \displaystyle \frac{x}{n a_{n}} \int_{0}^{\infty} x^{n} e^{-xt} dt$

$\displaystyle \Rightarrow f_{n+1} (x) = \displaystyle \frac{x}{n a_{n}} {a_{n+1}} f_{n+2} (x)$

Put $\displaystyle f_{n+1} (x) = \frac{a_{n+1}}{x^n}$ , and $\displaystyle f_{n+2} (x) = \frac{a_{n+2}}{x^{n+1}}$ to get :

$a_{n+2} = n a_{n}$

Now, At $x=1$ , our expression is $\displaystyle \frac{a_{20} a_{21}}{a_{18} a_{19}} = 18 \times 19 = 342$

We claim that

$f_n(x) = \frac{(n-2)!!}{x^{n-1}}$

for $n>1$ where $n!!$ is the double factorial. We proceed by induction.

We see that, $f_2(x) = \int_0^{\infty}e^{-tx}dt = \frac{1}{x}$

Assume that for $n=k>1$ ,

$f_k(x) = \frac{(k-2)!!}{x^{k-1}}$

Then, for $n=k+1$ , we have,

$f_{k+1}(x) = \int_0^{\infty}\frac{e^{-tx}}{f_k(t)}dt = \int_0^{\infty}\frac{t^{k-1}e^{-tx}}{(k-2)!!}dt$

and by repeated integration by parts,

$=\frac{1}{(k-2)!!}\left(\frac{-t^{k-1}e^{-tx}}{x} - \frac{(k-1)t^{k-2}e^{-tx}}{x^2} - \frac{(k-1)(k-2)t^{k-3}e^{-tx}}{x^3} + \dots - \frac{(k-1)!e^{-tx}}{x^k}\right)\big|_0^{\infty}$

Note that as $t\to\infty$ , each term goes to $0$ , and as $t\to0$ , each term except for the last also goes to $0$ . Thus the integral is equal to

$\frac{(k-1)!}{(k-2)!!x^k} = \frac{(k-1)!!}{x^k}$

Thus,

$\frac{f_{20}(1)f_{21}(1)}{f_{18}(1)f_{19}(1)} = \frac{18!!19!!}{16!!17!!} = \frac{19!}{17!} = 18*19 = \boxed{342}$