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So you have an integral like

$$ \int_{-\infty}^\infty{ \frac{dx}{1+4 x^2} } $$

Schaum's Calculus 5e recommends you write this as

$$ \lim_{a \to -\infty} \int_a^b{ \frac{dx}{1+4 x^2} } + \lim_{c \to \infty} \int_b^c{ \frac{dx}{1+4 x^2} } $$

Where b is chosen as a point where f(x) is defined.

Choosing b=0, you then get

$$ \lim_{a \to -\infty} \frac{1}{4} \int_a^0{ \frac{dx}{\frac{1}{4}+x^2} } + \lim_{c \to \infty} \frac{1}{4} \int_0^c{ \frac{dx}{\frac{1}{4}+x^2} } $$

$$ \lim_{a \to -\infty} \frac{1}{2} \tan^{-1}{ 2x } |_a^0 + \lim_{c \to \infty} \frac{1}{2} \tan^{-1}{ 2x } |_0^c $$

$$ 0 - \lim_{a \to -\infty} \frac{1}{2} \tan^{-1}{ 2a } + \lim_{c \to \infty} \frac{1}{2} \tan^{-1}{ 2c } - 0 $$

$$ = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} $$

Would it be correct to shortcut this as

  1. Take the indefinite integral with no limits $$ \frac{1}{4} \int{ \frac{dx}{ \frac{1}{4} + x^2} } = \frac{1}{2} \tan^{-1}{ 2x } $$

  2. Evaluate $$ = \frac{1}{2} \lim_{a \to -\infty} \lim_{b \to \infty} \tan^{-1}{ 2x } |_a^b $$

$$ = \frac{1}{2} \lim_{a \to -\infty} \lim_{b \to \infty} \left( \tan^{-1}{ 2b } - \tan^{-1}{ 2a } \right) $$

$$ = \frac{1}{2} \left( \frac{\pi}{2} + \frac{\pi}{2} \right) = \frac{ \pi }{2} $$

So my question is surrounding limit notation and evaluation. Is the above notation ok, or is it completely necessary to break it up into 2 limits?

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In this case it doesn't matter because the integral exists in a fairly strong sense. In general you will have to be careful because I think the answer can depend on the value of $b$. – Qiaochu Yuan May 2 '11 at 0:26
@Qiaochu: Hmmm... why? Replacing $b$ by $b'$ will add to the limit of the part from $a$ to $b$ the value of the integral from $b$ to $b'$ and will substract the same number from the limit of the integral from $b$ to $c$ so the results for $b$ and for $b'$ will coincide. – Did May 2 '11 at 8:04
@Didier: I may be misremembering - in particular, I may be confusing this with taking the limits on the left and right simultaneously - but I think there may be issues if the integrand has multiple singularities on the real line. – Qiaochu Yuan May 2 '11 at 14:57
@Qiaochu: I see. In the present case there are no singularities, nevertheless your point is one which is worth keeping in mind. Thanks. – Did May 2 '11 at 15:03
up vote 2 down vote accepted

To see why they have this approach, try

$$\int_{-\infty}^\infty{ \frac{8x \;dx}{1+4 x^2} } $$

using the recommended method [the indefinite integral is $\log(1+4 x^2) + \text{constant}$], and then using your suggestion, and alternatively what would happen if you had to work out

$$\lim_{c \to +\infty} \int_{-c}^c { \frac{8x \;dx}{1+4 x^2} }.$$

You would be looking with the recommended method at something like $\infty-\infty$, with your suggestion of doing the upper limit first at $\lim_{a \to -\infty} +\infty = +\infty$, and with the alternative method at $\lim_{c \to +\infty} 0 = 0$. In fact you want the alarm bells of the difference between two infinities.

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The integral can be interpreted in the Cauchy sense, though. – J. M. May 2 '11 at 2:03

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