Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?.

share|improve this question
    
I get an expression in terms of x and incomplete $\Gamma$ functions of imaginary arguments. –  Lucian May 7 at 12:57

1 Answer 1

You are right except that the lower limit of the outer integral should be $2$ and not $1$. That is:

$$I = \int_2^5 \int_1^{\sqrt{x-1}} \cos(x - \ln x) \, dy \, dx$$

share|improve this answer

Your Answer

 
discard

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.