Improper Integrals II. Improper Integrals II by Mika Seppälä Improper Integrals An integral is improper if either: the interval of integration is infinitely.

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Improper Integrals II

Improper Integrals II by Mika Seppälä Improper Integrals An integral is improper if either: the interval of integration is infinitely long or if the function f is either discontinuous or undefined at some points of the interval of integration (or both). Definition

Improper Integrals II by Mika Seppälä Improper Integrals Definition Assume that the function f is continuous on the interval [a, b). If the limit exists and is finite, the improper integral converges, and If the integral does not converge, then it diverges.

Improper Integrals II by Mika Seppälä Improper Integrals Example Assuming that 1 - p < 0. If 1 - p 0, the integral diverges. Assume that p 1.

Improper Integrals II by Mika Seppälä Improper Integrals Example Assuming that 1 - p > 0, i.e. p < 1. Assume that p > 0, and p 1. If p 0, the integral is not improper.

Improper Integrals II by Mika Seppälä Improper Integrals Example Assuming that a < 0. If a 0, the integral diverges. Assume that a 0.

Improper Integrals II by Mika Seppälä Basic Improper Integrals 1 1 converges if p > 1. 2 2 3 3 converges if p < 1. converges if a < 0.

Improper Integrals II by Mika Seppälä Basic Improper Integrals 4 4 converges if p < -1. 5 5 converges if p > -1. For all other values of the parameter the above integrals 1 - 5 diverge.

Improper Integrals II by Mika Seppälä Convergence of Improper Integrals Often it is not possible to compute the limit defining a given improper integral directly. In order to find out whether such an integral converges or not one can try to compare the integral to a known integral of which we know that it either converges or diverges.

Improper Integrals II by Mika Seppälä Convergence of Improper Integrals Consider the improper integral The graph of the function is shown in the figure.

Improper Integrals II by Mika Seppälä Idea of the Comparison Theorem The improper integral converges if the area under the red curve is finite.

Improper Integrals II by Mika Seppälä Idea of the Comparison Theorem We show this by finding a simple function, whose graph is the blue curve such that the area of the domain under the blue curve is finite. The improper integral converges if the area under the red curve is finite.

Improper Integrals II by Mika Seppälä Idea of the Comparison Theorem Observe that, for all x: Here the blue curve is the graph of and the red curve is that of

Improper Integrals II by Mika Seppälä Comparison Theorem Since and since we conclude that converges.

Improper Integrals II by Mika Seppälä Comparison Theorem Theorem A Let - a < b. Assume that 0 f(x) g(x) for all x, a < x < b. If the integral converges, then also converges, and

Improper Integrals II by Mika Seppälä Comparison Theorem Theorem B Let a < b. Assume that 0 f(x) g(x) for all x, a < x < b. If the integral diverges, then also diverges.

Improper Integrals II by Mika Seppälä Comparison Theorem Example Does the integral converge? We know that 0 < sin x < x, for 0 < x 1. Solution Hence for 0 < x 1.

Improper Integrals II by Mika Seppälä Comparison Theorem Example Does the integral converge? Solution (contd) Since for 0 < x 1, and since the integral diverges, also the integral diverges.

Improper Integrals II by Mika Seppälä Basic Improper Integrals 1 1 converges if p > 1. 2 2 3 3 converges if p < 1. converges if a < 0.

Improper Integrals II by Mika Seppälä Basic Improper Integrals 4 4 converges if p < -1. 5 5 converges if p > -1. For all other values of the parameter the above integrals 1 - 5 diverge. Use the Comparison Theorem to study the convergence (or divergence) of improper integrals by comparing them to an integral of one of the types 1 - 5.

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