An Elegant Rearrangement Of A Conditionally Convergent Iterated Integral

Introduction

$\backslash int\_0^1\backslash int\_0^1\; \backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash ,dy\backslash ,dx$

$\backslash int\_0^1\backslash int\_0^1$

How to evaluate this integral

$\backslash int\_0^1\backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash ,dy$

$y=x\backslash tan(\backslash theta),$

$dy=x\backslash sec^2(\backslash theta)\backslash ,d\backslash theta,$

$x^2+y^2=x^2+x^2\backslash tan^2(\backslash theta)=x^2(1+\backslash tan^2(\backslash theta))$

$0\backslash leq\; y\backslash leq\; 1,$

$0\backslash leq\; x\backslash tan(\backslash theta)\backslash leq\; 1,$

$0\backslash leq\backslash tan(\backslash theta)\backslash leq\; 1/x,$

$0\backslash leq\backslash theta\backslash leq\backslash arctan(1/x).$

$\backslash int\_0^\{\backslash arctan(1/x)\}$

$=\backslash frac\{1\}\{x\}\backslash int\_0^\{\backslash arctan(1/x)\}$

Now recall the trigonometric identities

$\backslash sin(\backslash arctan(1/x))=\backslash frac\{x\}\{\backslash sqrt\{1+x^2\}\}$

\ \mbox{and}\ \cos(\arctan(1/x))=\frac{1}{\sqrt{1+x^2}}. The expression above then becomes

$\backslash frac\{1\}\{1+x^2\}.$

This takes care of the "inside" integral with respect to y; now we do the "outside" integral with respect to x:

$\backslash int\_0^1\backslash frac\{1\}\{1+x^2\}\backslash ,dx$

=\left\arctan(x)\right_0^1 =\arctan(1)-\arctan(0)=\frac{\pi}{4}. Thus we have

$\backslash int\_0^1\backslash int\_0^1\backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash ,dy\backslash ,dx=\backslash frac\{\backslash pi\}\{4\}$

and

$\backslash int\_0^1\backslash int\_0^1\backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash ,dx\backslash ,dy=\backslash frac\{-\backslash pi\}\{4\}.$

The moral of the story

When

$\backslash int\_a^b\backslash int\_c^d\; \backslash left|f(x,y)\backslash right|\backslash ,dy\backslash ,dx=\backslash infty$

then the two iterated integrals

$\backslash int\_a^b\backslash int\_c^d\; f(x,y)\backslash ,dy\backslash ,dx\backslash \; \backslash mbox\{and\}\backslash \; \backslash int\_c^d\backslash int\_a^b\; f(x,y)\backslash ,dx\backslash ,dy$

may have different finite values. Elegant rearrangement of a conditionally convergent iterated integral, An

 

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