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First step of solving this improper integral.

$$\int_{-\infty}^{+\infty}\frac{dx}{x^2+1} = \lim_{a\mapsto\infty}\int_{-a}^{a}\frac{dx}{x^2+1}$$

I believe the correct way to proceed is:

$$\lim_{a\mapsto-\infty}\int_{a}^{0}\frac{dx}{x^2+1} +\lim_{b\mapsto+\infty}\int_{0}^{b}\frac{dx}{x^2+1}$$

But why is the first one incorrect?

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    The equality $\displaystyle\int_{-\infty}^{+\infty}f\,dx = \lim_{a\to\infty}\int_{-a}^{a}f\,dx$ is not true in general. That's not how the LHS is defined normally. The RHS is called the [principal value](https://en.wikipedia.org/wiki/Cauchy_principal_value). The way you believe is correct is indeed correct *by definition* of what is meant by the symbolization $\displaystyle\int_{-\infty}^{+\infty}\frac{dx}{x^2+1}$. – 2017-01-17
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    As long as your correct reading gives a sensible answer, the first will give the same answer. There are cases where the first will give what looks like a sensible answer but neither of the limits in the second exists independently. – 2017-01-17
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    Op, $\int_{-\infty}^{+\infty}f = \lim_{(m,n)\to(-\infty,+\infty)}\int_m^n f$. – 2017-01-17

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Because $\int_{-\infty}^{+\infty}\frac{dx}{x^2+1}$ exists, what you did first and second gives correct answers.

But if it is the case that $\int_{-\infty}^{\infty} f(x) dx$ does not exist then the following may or may not be true,

$\int_{-\infty}^{\infty} f(x) dx = \lim_{a \to \infty} \int_{-a}^{a} f(x) dx$

To see why, consider for example,

$$f(x)=x$$

Consider using the interval $[-t,2t]$ with $t \to \infty$ as bounds. Then use the interval $[-t,t]$ with $t \to \infty$ as bounds. Both the intervals approach $(-\infty,\infty)$ but..

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    The second sentence needs rewording to something like "If $\int_{-\infty}^\infty f(x)\,\mathrm dx$ does not exist, then $\lim_{a\to\infty}\int_{-a}^af(x)\,\mathrm dx$ may or may not exist". – 2017-01-17