Integral of xcos(xy) over a rectangle

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SUMMARY

This discussion focuses on evaluating double integrals over the rectangle defined by 1≤x≤2 and 0≤y≤1. The first integral, \(\int\int_{A}\frac{1}{x+y}dxdy\), was partially solved using the substitution \(\alpha(y)=\int^{2}_{1}\frac{1}{x+y}dx=ln(\frac{2+y}{1+y})\), but the user encountered difficulties in completing the evaluation. The second integral, \(\int\int_{A}xcos(xy)dxdy\), involved integration by parts, yet the user struggled with integrals such as \(\int\frac{sin(y)}{y}dy\) and \(\int\frac{cos(2y)}{y^{2}}dy\), which lack elementary primitives. Fubini's theorem allows for switching the order of integration, which may aid in solving these integrals.

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carlosbgois
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Hi there. I'm having some trouble with some double integrals here. All of them are to be evaluated on the rectangle_{} 1≤x≤2, 0≤y≤1, and the functions are:

1: \int\int_{A}\frac{1}{x+y}dxdy.
On this one I made α(y)=\int^{2}_{1}\frac{1}{x+y}dx=ln(\frac{2+y}{1+y}), and finally I should evaluate \int^{1}_{0}α(y)dy, and this is where I got stuck.

2: \int\int_{A}xcos(xy)dxdy.
I again have evaluated α(y), using integration by parts, but then I got stuck with some integrals like \int\frac{sin(y)}{y}dy, \int\frac{cos(2y)}{y^{2}}dy, and so on, which I couldn't find the primitive.

Any tips on how to evaluate those?
 
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You are fully allowed, by Fubini's theorem, to switch the order of integration.
:smile:
 
That's it, for both cases. Many thanks.
 

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