Proving convergence of integral

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SUMMARY

The discussion centers on proving the convergence of the double integral ##\int_0^1 \int_0^1 \frac{1}{1-xy}\, dx \, dy##. The user successfully demonstrates that ##\frac{1}{1 - xy}## can be expressed as the series ##\sum_{n=1}^\infty x^ny^n##, leading to the conclusion that the integral converges to ##\sum_{n=1}^\infty\frac{1}{(n+1)^2}##. The direct comparison test is applied to confirm convergence, as ##\frac{1}{(n+1)^2} < \frac{1}{n^2}##, which is a known convergent p-series. The user also notes the importance of ensuring the series is convergent before interchanging the sum and integral.

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  • Knowledge of the direct comparison test for series
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Homework Statement


Prove the following double integral is convergent.

##\int_0^1 \int_0^1 \frac{1}{1-xy}\, dx \, dy##

The Attempt at a Solution


This was a bonus question on my final exam in calc 3 yesterday, I just want to show my steps and see if they were right.

So I realized that
##\frac{1}{1 - xy} = \sum_{n=1}^\infty x^ny^n##

So
##\int_0^1 \int_0^1 \sum_{n=1}^\infty x^ny^n\, dx \, dy = \sum_{n=1}^\infty\frac{1}{(n+1)^2}##
Then I used the direct comparison test to show that

##\frac{1}{(n+1)^2} \lt \frac{1}{n^2}##
So since it is smaller than a convergent p-series, it is also convergent.
 
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Thanks for sharing this problem.
My only note would be that (I think) a series needs to be convergent in order to pass it through the integral. I would first take the partial sum to some large (finite) M, then you can pass it through and take the limit of the partial sums as M goes to infinity.

Aside from that, your work looks good. Nice and clean.
 

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