Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to integrate this:

$$ \int_{-1}^0 \int_{-2x-2}^{2x+2} x^2 y^2 + \sin(xy) e^{x^2} y^2 \;\operatorname d\!y \operatorname d\!x $$

I started integrating by parts,

$$ \int_{-1}^0 e^{x^2} \left({-y^2 \cos(xy)\over x} - \int_{-2x-2}^{2x+2} {-2y \cos(xy)\over x} \operatorname d\!y \right) \operatorname d\!x $$

but it seems to go on and on and on. It stops after two, but I have to do the same for $x$. Also, I have this other integral:

$$ \int_0^1 \int_{2x-2}^{-2x+2} x^2 y^2 + \sin(xy) e^{x^2} y^2 \;\operatorname d\!y \operatorname d\!x $$

I did it with Maple. It took 1-2 seconds and I got $4 \over 45$. It usually takes way longer even if it's a bit complicated.

Is there any trick I could use? Did I miss anything?

share|cite|improve this question
You may have already been aware of this, but be careful when using integration by parts with definite integrals. There shouldn't be a $y$ outside of your $\mathrm dy$ integral on your second line. – Mark S. Feb 16 '14 at 1:31
Can you please explain? $u \operatorname d\!v = u v - \int v \operatorname d\!u$, right? – ChrisVolkoff Feb 16 '14 at 2:07
That fact tells you about indefinite integrals. But you have to remember to plug in the limits of integration to the $uv$ part when you have a definite integral. In other words, $\int_a^bu\mathrm dv=uv\mid_a^b-\int_a^bv\mathrm du$. In particular, your $-y^2\cos(xy)/x$ term should have something like $\mid_{y=-2x-2}^{y=2x+2}$ next to it so you don't forget to plug in. (Separately, a variable outside of an integral with respect to that variable should be a red flag; at best it means you're using a variable in two different ways, but it usually means you've made an error.) – Mark S. Feb 16 '14 at 2:21
I totally forgot about evaluating $uv$! Thanks for pointing it out, I'll try to remember. – ChrisVolkoff Feb 16 '14 at 2:39
up vote 4 down vote accepted

The "difficult" part of your integrand, namely $\sin(xy) e^{x^{2}} y^{2}$, is an odd function integrated over a symmetric interval.

share|cite|improve this answer
Oh my, of course! $$ \int_{-2x-2}^{2x+2} \sin(xy) e^{x^2} y^2 \;\operatorname d\!y = 0 $$ Thank you very much, this simplifies everything! – ChrisVolkoff Feb 16 '14 at 0:09
You're very welcome. – Andrew D. Hwang Feb 16 '14 at 13:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.