Why do improper integrals split at different points give the same sum?

[DivGradCurl]

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

 

[NateTG]

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.

 

[DivGradCurl]

That sounds about right. I mean, I can't see anything else that could be done.