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Are there any special cases that make the following true

$$\int\frac{f(x)}{g(x)} dx = \frac{\int f(x)\ dx}{\int g(x) \ dx}$$

Thanks

Tyrone
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  • did $f(x)=g(x)$ is a special case? – cand Feb 02 '15 at 17:10
  • At least, in some sense, you should have the measure of the set over which you integrate to be one, since if you average it out, in the second one they cancel. – ploosu2 Feb 02 '15 at 17:11
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    Of cource, $f(x) = 0$ is a special case althoug rather trivial, where this is true. – ploosu2 Feb 02 '15 at 17:13
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    Not anything which can be formulated in terms of mathematics is a good question in the sense that answering it will provide insights. – Fabian Feb 02 '15 at 17:14
  • @Fabian what do you mean? – Tyrone Feb 02 '15 at 17:15
  • @Fabian I think this is a question worthi considering. Think about the product rule. The integral doesn't go directly to products but we have the product rule. – ploosu2 Feb 02 '15 at 17:16
  • @Tyrone Now, now, according to the rules of this place, it's good practise to give background information and what you have tried about the problem. – ploosu2 Feb 02 '15 at 17:19
  • @Fabian I agree but the way you are asking is implying that the question doesnt make sense. – Tyrone Feb 02 '15 at 17:20
  • @Tyrone: the first comment was rather general. If you read it carefully, you understand that the problem is exactly not that it does not make sense (your question makes perfectly sense). With the second, I did imply that I would be surprised if this question did come up in some application but again not that the question does not make any sense. By the way, I am not one of the downvoters. – Fabian Feb 02 '15 at 17:22
  • @Fabian alright sure. – Tyrone Feb 02 '15 at 17:23
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    Without proper context and/or at least some self work shown, this looks like a whimsical question shot out of the blue. I can understand why people would want to know more about this question's background: at least, to get some insight from answering it or from dealing with it. BTW, $;g(x)=f(x);$ doesn't make the cut. – Timbuc Feb 02 '15 at 17:27
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    Definite or indefinite integral? There is a fundamental ambiguity here because on the right, you have two integration constants, one in the denominator. – orion Feb 02 '15 at 18:03

1 Answers1

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Letting $h=f/g$, the question is equivalent to $$ \int h\int g=\int g\,h. $$ If $a\,b=a+b$ then $$ \int e^{ax}\,dx\int e^{bx}\,dx=\int e^{ax}e^{bx}\,dx $$ for an for appropriate choice of the integration constants.

Julián Aguirre
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