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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

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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: $$\frac{1}{2}\int_{u=-1}^0\int_{v=1}^{2+u}\cos\left(u\right)e^{v}dvdu +\frac{1}{2}\int_{u=0}^2\int_{v=1}^{2-u}\cos\left(u\right)e^{v}dvdu \\+\frac{1}{2}\int_{u=2}^3\int_{v=1}^{u-2}\cos\left(u\right)e^{v}dvdu.$$

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  • $\begingroup$ @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
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Hint

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.

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  • $\begingroup$ 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

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