A calculus problem by Atul Kumar Ashish

Changing variables, with $X = x+y$ , we see that $\int_0^\infty \int_0^\infty \frac{e^{-(x+y)}}{x+y}\,dx\,dy \; = \; \int_0^\infty\,dX \int_0^X \frac{e^{-X}}{X}\,dx \; = \; \int_0^\infty e^{-X}\,dX \; = \; 1$

This is what called the Feynman’s method of parametrization. Define a integral for $a>0$ by :

$\displaystyle I(a) = \int_0^\infty \int_0^\infty e^{-a(x+y)}{x+y}\; dx \; dy$

Now we want to differentiate it with respect to ‘a’ , however there are strict conditions over which this differentiation is valid but we notice that function is always positive

and never negative in the given intervals which eases things a bit. The differentiated integral is also convergent and integrable, so we differentiate to get,

$\displaystyle \dfrac{dI}{da} = -\int_0^\infty \int_0^\infty \dfrac{d}{da}\left(\dfrac{e^{-a(x+y)}}{x+y} \right) \; dx \; dy =- \int_0^\infty \int_0^\infty e^{-a(x+y)}\; dx\; dy$

Now the integral is in variable separable form and we can evaluate that to get,

$\displaystyle \dfrac{dI}{da} = -\dfrac{1}{a^2}$

Now what ? Yes you are correct , just integrate both sides from $\infty$ to $1$ this will give $I(1)-\lim_{t\to\infty}I(t) = 1$ . Notice that $\lim_{t\to\infty}I(t)=0$ as $\lim_{t\to\infty}e^{-tu}=0$ This gives us:

$\displaystyle \int_0^\infty \int_0^\infty \dfrac{e^{-(x+y)}}{x+y}\; dx \;dy = \boxed{1}$