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Assume the following function, with $x, a, b \in \! \mathbb{R}$

$$ f(x,a,b) = \begin{cases} x+a+b & \mbox{for } ~ a-b \le x \le a+b \\ 0 & \mbox{elsewhere} \end{cases} $$

How can the following integral be computed?

$$ I(x) = \int_{-\infty}^{+\infty} \! \int_{-\infty}^{+\infty} \! f(x,a,b) \, \mathrm{d}a ~\mathrm{d}b $$

thekyria
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1 Answers1

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The domain of integration can be changed since $f(x,a,b) = 0$ unless $(a,b) \in (-\infty,x]\times[x,\infty)$. The result is: $$I(x) = \int_x^\infty \int_{-\infty}^x (x+a+b) da \, db$$

However, this integral does not converge.

Update (original question was edited): The function is non-zero only if $a \in [x-b,x+b]$, so we can write: $$I(x) = \int_{-\infty}^\infty \int_{x-b}^{x+b} (x+a+b) da \, db$$

This integral still does not converge.

Josh B
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    And, more generally: an integral of a function defined in cases can be done by converting it to several integrals, one for each case. Here, all other cases are zero, so Harry left them out. A typical example of this is an integral involving absolute values. – GEdgar Sep 19 '14 at 13:55
  • For example, the first integral begins as $\int_{-\infty}^{\infty} f , da$, but we can re-write this as $\int_{-\infty}^{x} f , da + \int_{x}^{\infty} f , da$ and the second integral is 0. – Josh B Sep 19 '14 at 14:08
  • Will edit my original answer now to reflect the change in question. – Josh B Sep 19 '14 at 14:19