How to change the order of integration: ∫ − 1 1 d x ∫ 1 − x 2 2 − x 2 f ( x , y ) d y

Question

How to change the order of integration:      ∫          −      1        1    d  x      ∫          1      −              x        2                    2      −              x        2              f  (  x  ,  y  )  d  y

Layne Bailey · Accepted Answer

The easy way to do this kind of things is to rewrite integral boundaries into indicator functions:  I  :=      ∫          −      1        1    d  x      ∫          1      −              x        2                    2      −              x        2              d  y    f  (  x  ,  y  )  =      ∫          −      1        1    d  x      ∫          0              2        d  y    f  (  x  ,  y  )      χ          1      −              x        2            &amp;lt;      y      &amp;lt;      2      −              x        2            where       χ    condition   is 1 if the condition is met and 0 otherwise. Then you can switch easily  I  =      ∫          0              2        d  y      ∫          −      1        1    d  x    f  (  x  ,  y  )      χ          1      −              x        2            &amp;lt;      y      &amp;lt;      2      −              x        2              =      ∫          0              2        d  y      ∫          −      1        1    d  x    f  (  x  ,  y  )      χ          1      −      y      &amp;lt;              x        2            &amp;lt;      2      −      y      Given that the lower bound   1  −  y for       x    2   changes sign at   y  =  1, it makes sense to cut the integral on y at   y  =  1  I  =      ∫          0              1        d  y      ∫          −      1        1    d  x    f  (  x  ,  y  )      χ          1      −      y      &amp;lt;              x        2              +      ∫          1              2        d  y      ∫          −      1        1    d  x    f  (  x  ,  y  )      χ                  x        2            &amp;lt;      2      −      y      Notice how one could drop the condition       x    2    &amp;lt;  2  −  y in the first integral, and   1  −  y  &amp;lt;      x    2   in the second, as these are now always true.In the second integral, the condition is equivalent to   −      2    −    y    &amp;lt;  x  &amp;lt;      2    −    y  , and these bounds are within   [  −  1  ,  1  ], so things are easy: we can replace the indicator function with the new bounds. In the first integral the condition is equivalent to   x  &amp;gt;      1    −    y   or   x  &amp;lt;  −      1    −    y  . This splits the first integral into two parts:  I  =      ∫          0              1        d  y      ∫          −      1              −              1        −        y              d  x    f  (  x  ,  y  )  +      ∫          0              1        d  y      ∫                  1        −        y              1    d  x    f  (  x  ,  y  )  +      ∫          1              2        d  y      ∫          −              2        −        y                            2        −        y              d  x    f  (  x  ,  y  )  .This is the relation you give.

How to change the order of integration: <msubsup> ∫ <mrow class="MJX-Te

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