Plot of 1/(sqrt(x)*(x+1)) from 0.093 to 3.0

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A “proper” Riemann integral assumes the integrand is defined and finite on a closed and bounded interval, bracketed by the limits of integration. An improper integral occurs when one or more of these conditions is not satisfied. In some cases such integrals may be defined by considering the limit of a sequence of proper Riemann integrals on progressively larger intervals.

If the interval is unbounded, for instance at its upper end, then the improper integral is the limit as that endpoint goes to infinity.

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

If the integrand is only defined or finite on a half-open interval, for instance (a,b], then again a limit may provide a finite result.

\int_{a}^{b} f(x)dx = \lim_{\epsilon \to 0} \int_{a+\epsilon}^{b} f(x)dx

That is, the improper integral is the limit of proper integrals as one endpoint of the interval of integration approaches either a specified real number, or ∞, or −∞. In more complicated cases, limits are required at both endpoints, or at interior points.

Consider, for example, the function \tfrac{1}{(x+1)\sqrt{x}} integrated from 0 to ∞ (shown right). At the lower bound, as x goes to 0 the function goes to ∞, and the upper bound is itself ∞, though the function goes to 0. Thus this is a doubly improper integral. Integrated, say, from 1 to 3, an ordinary Riemann sum suffices to produce a result of \tfrac{\pi}{6}. To integrate from 1 to ∞, a Riemann sum is not possible. However, any finite upper bound, say t (with t > 1), gives a well-defined result, \tfrac{\pi}{2} - 2\arctan \tfrac{1}{\sqrt{t}}. This has a finite limit as t goes to infinity, namely \tfrac{\pi}{2}. Similarly, the integral from 13 to 1 allows a Riemann sum as well, coincidentally again producing \tfrac{\pi}{6}. Replacing 13 by an arbitrary positive value s (with s < 1) is equally safe, giving -\tfrac{\pi}{2} + 2\arctan\tfrac{1}{\sqrt{s}}. This, too, has a finite limit as s goes to zero, namely \tfrac{\pi}{2}. Combining the limits of the two fragments, the result of this improper integral is

 \int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x}} &{} = \lim_{s \to 0} \int_{s}^{1} \frac{dx}{(x+1)\sqrt{x}}
   + \lim_{t \to \infty} \int_{1}^{t} \frac{dx}{(x+1)\sqrt{x}} \\
  &{} = \lim_{s \to 0} \left( - \frac{\pi}{2} + 2 \arctan\frac{1}{\sqrt{s}} \right)
   + \lim_{t \to \infty} \left( \frac{\pi}{2} - 2 \arctan\frac{1}{\sqrt{t}} \right) \\
  &{} = \frac{\pi}{2} + \frac{\pi}{2} \\
  &{} = \pi .

This process does not guarantee success; a limit may fail to exist, or may be unbounded. For example, over the bounded interval 0 to 1 the integral of \tfrac{1}{x} does not converge; and over the unbounded interval 1 to ∞ the integral of \tfrac{1}{\sqrt{x}} does not converge.

The improper integral
\int_{-1}^{1} \frac{dx}{\sqrt[3]{x^2}} = 6
is unbounded internally, but both left and right limits exist.

It may also happen that an integrand is unbounded at an interior point, in which case the integral must be split at that point, and the limit integrals on both sides must exist and must be bounded. Thus

 \int_{-1}^{1} \frac{dx}{\sqrt[3]{x^2}} &{} = \lim_{s \to 0} \int_{-1}^{-s} \frac{dx}{\sqrt[3]{x^2}}
   + \lim_{t \to 0} \int_{t}^{1} \frac{dx}{\sqrt[3]{x^2}} \\
  &{} = \lim_{s \to 0} 3(1-\sqrt[3]{s}) + \lim_{t \to 0} 3(1-\sqrt[3]{t}) \\
  &{} = 3 + 3 \\
  &{} = 6.

But the similar integral

 \int_{-1}^{1} \frac{dx}{x} \,\!

cannot be assigned a value in this way, as the integrals above and below zero do not independently converge. (However, see Cauchy principal value.)



Welcome to my blog. My name is Nico. Admin of this blog. I am a student majoring in mathematics who dreams of becoming a professor of mathematics. I live in Kwadungan, Ngawi, East Java. Hopefully in all the posts I can make a good learning material to the intellectual life of the nation. After the read, leave a comment. I always accept criticism suggestion to build a better me again .. Thanks for visiting .. : mrgreen:

Posted on September 2, 2011, in education and tagged , , , , , , , . Bookmark the permalink. 2 Comments.

  1. Pusing aku belajar matematika bong.

  2. I hate math. LOL I tried to be a math major and had no clue what I was in for. But I love numbers and actually, I love math and I hope to learn some stuff here.




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