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For instance, is defining $$\int_{-a}^af(x)dx=\int_{-a}^0f(t)dt+\int_0^af(x)dx$$ an ok thing to do? Thanks.

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Yes, this is allowed. In fact, the variable itself is irrelevant since $\displaystyle \int_a^b f(x)\,dx=\int_a^b f(t)\,dt=\int_a^b f(\ddot\smile)\,d\,\ddot\smile$. –  Brian Fitzpatrick Apr 19 '13 at 15:09
    
lol, thank you! –  ohad Apr 19 '13 at 15:10
    
@chad : you don't have to "define" it, it's true. And $f$ doesn't have to be continuous either, if you are talking about a regular Riemann or Lebesgue integral. –  Stefan Smith Apr 19 '13 at 21:35

4 Answers 4

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Yes, you can do this, as long as $f$ is a measurable function, if we are talking about Lebesgue integral, or just a Riemann-Integrable function in the Riemann case.

However you have to be careful when integrating this way, since

$$\lim\limits_{a\rightarrow\infty}\int_{-a}^af(x)dx\neq \int _{-\infty }^\infty f(x)dx$$ unless the second exists.

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You shouldn't define the integtral like so. You should prove it's true (and it is).

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Yes, the variable you integrate over is a "dummy variable". Each of the three integrals you have represents a number. You can think of it as a $u$ substitution with $t=x$ if you want.

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Yes, this is valid. As long as the function is continuous, you are OK. But you have to define the relationship of x to t, and make sure that the function is defined throughout.

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