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Solving for C when we have $C\int_0^\infty \int_0^\infty \frac{e^\frac{-(x_1+x_2)}{2}}{x_1+x_2} \,dx_1 \,dx_2=1$

$$\int_0^\infty \int_0^\infty \frac{e^\frac{-(x_1+x_2)}{2}}{x_1+x_2} \,dx_1 \,dx_2$$

Let $u=x_1+x_2$ and $\,du=\,dx_1$.

$$\int_0^\infty \int_{x_2}^\infty \frac{e^\frac{-u}{2}}{u} \,du \,dx_2 $$

How do I compute the inner integral?

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From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrd\theta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $\theta$ integral of $\sin(\theta)\cos(\theta)$ \begin{align} 4\iint\frac{\exp(-r^2/2)}{r^2}r^3\sin(\theta)\cos(\theta)drd\theta &= 4\iint\exp(-u)\sin(\theta)\cos(\theta)dud\theta\\ &= -4\int\exp(-u)\Bigl|_0^{\infty}\sin(\theta)\cos(\theta)d\theta\\ &=\cdots \end{align}


If you obtained $\frac{1}{16\pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.

When we convert to $r$ and $\theta$, the bounds become $0<r<\infty$ and $0<\theta<\pi/2$

Mouse over for solution:

\begin{align}&= 4\int_0^{\pi/2}\sin(\theta)\cos(\theta)d\theta\\&=2\int_0^{\pi/2}\sin(2\theta)d\theta\\&= -\cos(2\theta)\Bigr|_0^{\pi/2}\\&=1 + 1\\&=2\end{align}

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  • $\begingroup$ Perfect so $c=\frac{1}{16\pi^2}$ $\endgroup$ – Username Unknown Nov 21 '14 at 3:37
  • $\begingroup$ @UsernameUnknown You should get $C=1/2$ $\endgroup$ – dustin Nov 21 '14 at 3:37
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Integration by parts:)) $$ \int_0^{\infty}\int_{x_2}^{\infty}\frac{e^{-\frac u2}}udu\, dx_2=\int_0^{\infty}x_2\frac{e^{-\frac {x_2}2}}{x_2}dx_2=2. $$

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