How do you get from$$\int^\infty_0\int^\infty_0e^{-(x+y)^2} dx\ dy$$to
$$\frac{1}{2}\int^\infty_0\int^u_{-u}e^{-u^2} dv\ du?$$ I have tried using a change of variables formula but to no avail.
Edit: Ok as suggested I set $u=x+y$ and $v=x-y$, so I can see this gives $dx dy=\frac{1}{2}dudv$ but I still can't see how to get the new integration limits. Sorry if I'm being slow.
Wednesday, 25 February 2015
calculus - How to Evaluate $int^infty_0int^infty_0e^{-(x+y)^2} dx dy$
Subscribe to:
Post Comments (Atom)
real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
$$ 3x+6y+5z=7 $$ The general solution to this linear Diophantine equation is as described here (Page 7-8) is: $$ x = 5k+2l+14 $$ $$ y = -l $...
-
How to show the following inequality in Measure Theory: If $f$ is a non-negative measurable function defined on a measurable set $E$ then ...
-
I need help to compute the following integral: $$\int_{-\infty}^{\infty}\frac{z^4}{1+z^8}dz$$ I need to use Cauchy's residue theorem. I ...
No comments:
Post a Comment