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Examples: Inproper Integral

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z overview
Consider the integral

$\displaystyle \int_{-\infty}^{\infty} \frac{1+2x}{1+x^2}\,dx\,. $

An antiderivative of the integrand is

$\displaystyle \arctan(x)+\ln(1+x^2)\,. $

The wrong approach

$\displaystyle \int_{-\infty}^{\infty} \frac{1+2x}{1+x^2}\,dx$ $\displaystyle = \lim_{b\to\infty} \int_{-b}^{b} \frac{1+2x}{1+x^2}\,dx = \lim_{b\to\infty} \left[\arctan(x)+\ln(1+x^2)\right]_{-b}^{b}$    
  $\displaystyle = \lim_{b\to\infty} \left(2\arctan(b)\right) = \pi$    

results $ \pi$ for the inproper integral. However, by the definition of the inproper integral one have to examine the upper and lower limit separately. As neither the limit

$\displaystyle \lim_{c\to -\infty} \int_{c}^{0} \frac{1+2x}{1+x^2}\,dx$ $\displaystyle = \lim_{c\to -\infty} \left( -\arctan(c) - \ln(1+c^2) \right)$    

nor the limit


$\displaystyle \lim_{d\to \infty} \int_{0}^{d} \frac{1+2x}{1+x^2}\,dx$ $\displaystyle = \lim_{d\to\infty} \left( \arctan(d) + \ln(1+d^2) \right)$    

exists, also the inproper integral

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

not exists.

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  automatisch erstellt am 12.  9. 2008