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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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    @Qiaochu: I see. In the present case there are no singularities, nevertheless your point is one which is worth keeping in mind. Thanks.2011-05-02

1 Answers 1

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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.2011-05-02