Prove Integrals Equality for Real a & b

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To prove the equality of integrals for real numbers a and b, one can utilize the property of splitting integrals at arbitrary points. The integral from negative infinity to a can be expressed as the sum of the integral from negative infinity to b and the integral from b to a. This leads to the formulation that combines both sides of the equation, demonstrating that the two expressions are equal. The discussion emphasizes that this approach is fundamentally algebraic and relies on the convergence of the integral. Overall, the method of splitting the integral at points a and b effectively illustrates the equality.
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If \int _{-\infty} ^{\infty}f(x)\: dx is convergent and a and b are real numbers, show that

\int _{-\infty} ^a f(x)\: dx + \int _a ^{\infty}f(x)\: dx = \int _{-\infty} ^b f(x)\: dx + \int _b ^{\infty}f(x)\: dx


I'm clueless on how to show it other than by drawing what is stated: a generic finite integral being split into two finite pieces for each arbitrary point. Is there any other way to approach this problem?

Thanks
 
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thiago_j said:
I'm clueless on how to show it other than by drawing what is stated: a generic finite integral being split into two finite pieces for each arbitrary point. Is there any other way to approach this problem?

Perhaps you could use something like:
\int _{-\infty} ^a f(x)\: dx + \int _a ^{\infty}f(x)\: dx = \int _{-\infty} ^a f(x)\: dx + \int _{a} ^b f(x)\: dx +\int _b ^{\infty}f(x)\: dx = \int _{-\infty} ^b f(x)\: dx + \int _b ^{\infty}f(x)\: dx

But that's really just an algebraic representation of what you're suggesting.
 
That sounds about right. I mean, I can't see anything else that could be done.
 
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