Integral Series

First we let $\int_0^{\infty} f(x) dx = R = 63 \pi$ .

As stated in the problem, the fact that the expression

$\large \frac{\displaystyle\int_{n\alpha}^{(n+1)\alpha} \! f(x) \mathrm{d}x }{ \displaystyle \int_{(n-1)\alpha}^{n\alpha} f(x) \mathrm{d}x }$

is constant for any integer $n \geq 1$ given any positive real number $\alpha$ suggests that when $R$ is subdivided along the x-axis into partitions of length $\alpha$ , the ratio of consecutive resulting sub areas is constant. This can be interpreted that these sub areas represent a geometric sequence with a certain common ratio $k$ , whose sum equals the region $R = 63 \pi$ .

We can choose any arbitrary $\alpha$ for that matter, as the common ratio $k$ will be dependent on it. Setting $\alpha=1$ eases the task of finding the first sub-area, as we already have the value of $\int_0^1 f(x) dx$ , which is $7\pi$ , which we will label as $A$ .

To solve for $k$ ,

$\int_0^{\infty} f(x) dx = \sum\limits_{n=0}^{\infty} \int_n^{n+1} f(x) dx$

$63 \pi = \frac{7 \pi}{1-k}$

And we get $k= \frac{8}{9}$ .

Now, we know that the antiderivative of $f(x)$ is $F(x) + C$ for some arbitrary constant C. By the Fundamental Theorem of Calculus, we see that

$\int_{0}^1 f(x) dx = F(1)-F(0) = A$

If we assign a particular value $B$ for $F(0)$ , we see that $F(1) = A + B$ .

Working our way to $F(2)$ onwards, we get the following:

$F(2) = A(1+k) + B$

$F(3) = A(1+k+k^2) + B$

$F(4) = A(1+k+k^2+k^3) + B$

Now we get its pattern (of a finite geometric series, that is), such that we can formulate an explicit formula for $F(x)$ as

$F(x) = \frac { A( 1 - k^x)}{1-k} + B$

And we just take its derivative and we'll have $f(x)$ !

$f(x) = \frac{d}{dx} [ \frac{A}{1-k} (1 - k^x ) + B ]$

$f(x) = \frac{ -Ak^x \ln k }{1-k}$

$f(x) = \frac{ Ak^x \ln k}{k-1}$

And so we can readily get $f(0)$

$f(0) = \frac { 7 \pi \ln \frac {8}{9}}{ - \frac{1}{9}}$

$f(0) = 63 \pi ( \ln 9 - \ln 8) \approx 23.311658$

So $\lfloor 1000 \times f(0) \rfloor = 23311$