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Is there anything I can do with a double integral of the following form?

$$ \int_0^1 \left(\int_0^x f(j) dj\right)^\alpha g(x) dx$$

For $\alpha = 1$, I can do

$$ \int_0^1 \int_0^x f(j) dj g(x) dx = \int_0^x f(j) \int_0^1 g(x) dx dj$$,

where the inner integral has a well defined solution.

For general $\alpha$, I know that there is a difference between $x_1^a + x_2^a$ and $(x_1 + x_2)^a$, but is there perhaps a trick I can use to still change the order of integration?

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    $\begingroup$ It is not true that your integral equals $\int_0^x f(j) \int _i^1 g(x) dx dj$. This would yield something that depends on $x$, instead of a constant. You have to rewrite your integration domain ($\{(x,j)| i\le x \le 1, 0\le y \le x \}$) so that the bounds for $y$ are constant. This is not possible here unless $i=0$ (in this case you could write the domain as $\{(x,j)| y\le x \le 1, 0\le y \le 1 \}$). $\endgroup$
    – Kuifje
    Jan 17, 2017 at 17:58
  • $\begingroup$ @Kuifje When you say $y$ you mean $j$ right? $\endgroup$
    – FooBar
    Jan 17, 2017 at 18:04
  • $\begingroup$ In spirit of the original question, I will change $i\to 0$. $\endgroup$
    – FooBar
    Jan 17, 2017 at 18:14
  • $\begingroup$ yes I meant $j$...thanks for spotting that. $\endgroup$
    – Kuifje
    Jan 17, 2017 at 18:20
  • $\begingroup$ With $h(x) = \int_0^x f(y){\rm d}y$ you are asking if there is possible to simplify $\int_0^1 h(x)^\alpha g(x) {\rm d}x$ just knowing that $h'(x) = f(x)$. It's just too general of an integral with to free functions to allow for some major simplification. You will have to evaluate it on a case by case basis. $\endgroup$
    – Winther
    Jan 17, 2017 at 18:21

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