# intint_D cosxy dxdy, where D: {[x>=0],[y>=0],[x>=y]}?

Jun 25, 2018

This integration is in the first quadrant, under the line $y = x$

$\int {\int}_{D} \cos x y \setminus \mathrm{dx} \setminus \mathrm{dy}$

$= \textcolor{b l u e}{{\int}_{x = 0}^{\infty} \mathrm{dx} \setminus {\int}_{y = 0}^{x} \mathrm{dy} q \quad \cos x y} \textcolor{g r e e n}{\equiv} \textcolor{red}{{\int}_{y = 0}^{\infty} \mathrm{dy} \setminus {\int}_{x = y}^{\infty} \mathrm{dx} q \quad \cos x y}$

Following the blue integration:

$= {\int}_{x = 0}^{\infty} \mathrm{dx} q \quad {\left[\frac{\sin x y}{x}\right]}_{y = 0}^{x}$

$= {\int}_{x = 0}^{\infty} \mathrm{dx} q \quad \frac{\sin {x}^{2}}{x}$

$z = {x}^{2} q \quad \mathrm{dz} = 2 x \setminus \mathrm{dx} = 2 \sqrt{z} \setminus \mathrm{dx}$

$= {\int}_{0}^{\infty} \mathrm{dz} \frac{1}{2 \sqrt{z}} q \quad \frac{\sin z}{\sqrt{z}}$

$= \frac{1}{2} {\underbrace{{\int}_{0}^{\infty} \mathrm{dz} q \quad \frac{\sin z}{z}}}_{\text{Feynmann Integration}} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$

For that last integration, a sketch:

Create function:

• $I \left(a\right) = {\int}_{0}^{\infty} \mathrm{dz} q \quad \frac{\sin z}{z} {e}^{- a z} q \quad \triangle q \quad a \ge 0$

Differentiate wrt $a$:

• $I ' \left(a\right) = {\int}_{0}^{\infty} \mathrm{dz} q \quad - z \frac{\sin z}{z} \setminus {e}^{- a z}$

$= - {\int}_{0}^{\infty} \mathrm{dz} q \quad \sin z \setminus {e}^{- a z}$

That, after 2 rounds of IBP, is:

$I ' \left(a\right) = - \frac{1}{{a}^{2} + 1}$

Integrate wrt $a$

$I \left(a\right) = - \arctan \left(a\right) + C q \quad \square$

Combining:

• $\triangle : q \quad {\lim}_{a \to \infty} {\int}_{0}^{\infty} \mathrm{dz} q \quad \frac{\sin z}{z} {e}^{- a z} = 0$

• $\square : q \quad I \left(a \to \infty\right) = - \arctan \left(a \to \infty\right) + C = 0$

$\implies C = \frac{\pi}{2}$

$\implies I \left(a\right) = - \arctan \left(a\right) + \frac{\pi}{2}$

And:

• $I \left(0\right) = {\int}_{0}^{\infty} \mathrm{dz} q \quad \frac{\sin z}{z} = \frac{\pi}{2}$