I need to evaluate the following integral :

$$\int_{ -1}^1\int_{1+x}^1\cos\left(x+y\right)e^{(y-x)}dydx$$

I know that I need to change the variables by using substitution $u = x+y$ and $v =y-x$ but I am confused about changing the limits of the new integral ,

I am trying to get the limits by drawing the graph of given limits in (x,y) and then , draw the corresponding graph in u-v plane using given equation,

But I am still not getting it .

Can please someone explain on how to change limits of this integral, and how should I proceed while tackling such problems of the same kind ?

Thank you


By changing the limits the given integral is equal to $$\frac{1}{2}\int_{u=-1}^1\int_{v=1}^{2-|u|}\cos\left(u\right)e^{v}dvdu+\frac{1}{2}\int_{u=1}^3\int_{v=1}^{|u-2|}\cos\left(u\right)e^{v}dvdu$$ where $1/2$ is due to the Jacobian of the transformation. Please check the new limits by making a drawing of the domain in the $uv$ plane and by comparing it with the domain in the $xy$ plane.

You can even further split the intervals of integration in order to eliminate the absolute values: $$\frac{1}{2}\int_{u=-1}^0\int_{v=1}^{2+u}\cos\left(u\right)e^{v}dvdu +\frac{1}{2}\int_{u=0}^2\int_{v=1}^{2-u}\cos\left(u\right)e^{v}dvdu \\+\frac{1}{2}\int_{u=2}^3\int_{v=1}^{u-2}\cos\left(u\right)e^{v}dvdu.$$

  • $\begingroup$ @sat091 Any further doubt? $\endgroup$ – Robert Z Aug 16 at 10:10
  • $\begingroup$ sorry for the late response, but I have one question, In the second integral where $v$ from $ |u-2|$ to $1$, the limits for $v$ should be reversed, From the graph I have drawn the line $v=1$ is above the two lines $v+u=2$ and $u-v = 2$,therefore I think the limits should be reversed, Can you please check this again ? thanks for the help. $\endgroup$ – sat091 Aug 16 at 20:11
  • $\begingroup$ @sat091 I think it is correct. Note that in the original integral $1+x\leq 1$ for $x\in [-1,1]$ but $1+x\geq 1$ for $x\in [0,1]$. $\endgroup$ – Robert Z Aug 16 at 20:22


In that specific case, you don’t need to change the variables. Just develop $\cos(x+y)$ using usual trigonometric formula and use $e^{y-x}=e^ye^{-x}$.

You then have to integrate maps with $x,y$ as separated variables.

  • $\begingroup$ Thanks for your answer, But this question came on my exam and I had to solve this by changing the limits of integral , Can you please explain how should I change the limits of given integral. $\endgroup$ – sat091 Aug 16 at 7:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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