Seeing Double

Relevant wiki: Integration Tricks

The given integral is equal to the double integral $\large \displaystyle\int_{0}^{\infty} \int_{\sqrt{2}}^{\pi} e^{-xy} dydx$ .

Changing the order of integration, the integral then becomes

$\large \displaystyle \int_{\sqrt{2}}^{\pi} \int_{0}^{\infty} e^{-xy} dxdy = \int_{\sqrt{2}}^{\pi} \dfrac{1}{y} dy = \ln(\pi) - \ln(\sqrt{2}) = 0.798156....$ ,

and so $\lfloor 1000S \rfloor = \boxed{798}$ .

Define:

$f(a) = \int_{0}^{\infty} \frac{e^{-ax} -1}{x} dx ; f(0)=0$

Differentiating with respect to a, we get

$f'(a) = \frac{-1}{a}$

Integrating both sides,

$f(\sqrt{2}) - f(\pi) = \int_{\pi}^{\sqrt{2}} f'(t) dt = \int_{\pi}^{\sqrt{2}} \frac{-1}{t} dt = ln(\frac{\pi}{\sqrt{2}})$

Hence, the value of the integral is $ln(\frac{\pi}{\sqrt{2}})$ .

We can solve this via Frullani's integral.

$\displaystyle\int_0^{\infty} \frac{ f(ax) - f(bx)}{x} \mathrm{d}x = [ f(0) - f(\infty) ] \ln \frac{b}{a}$

Setting $f(x) = e^{-x}$ , we see that $f(0=1$ , and $f(\infty) = 0$ . Also, based on the problem, $a= \sqrt{2}$ , and $b= \pi$ .

So,

$\displaystyle \int_0^{\infty} \frac{ e^{-\sqrt{2}}x - e^{-\pi x} }{x} \mathrm{d}x = [ 1-0] \ln \frac{\pi}{\sqrt{2}} \approx 0.78915629$

Hence, $\lfloor 1000A \rfloor = 789$ .