# double integral.

I just received this problems from a friend, and I think its a HW problem.

its:

$$\int_1^e \int_{1+y^2}^5 \cos (x- \ln x) \ dx \ dy$$

I looked at it, and If I did graph the the region right then the point where y=e intersects with is beyond where where x=5 intersects with $$\ x=1+y^2$$ so looks like the integral should be

$$\int_1^2 \int_{1+y^2}^5 \cos (x- \ln x) \ dx \ dy$$

I rewrote the integral as: $$\int_1^5 \int_1^{\sqrt {x-1}} \cos (x- \ln x) \ dy \ dx$$

is my reordering the region good?, and is there any suggestion on how to compute this integral?.

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I get an expression in terms of x and incomplete $\Gamma$ functions of imaginary arguments. – Lucian May 7 '14 at 12:57

You are right except that the lower limit of the outer integral should be $2$ and not $1$. That is:
$$I = \int_2^5 \int_1^{\sqrt{x-1}} \cos(x - \ln x) \, dy \, dx$$