What exact interchange does the DCT justify here?
Conditions
- The previously stated convergence, domination, and finite-integral conditions are assumed to hold.
- The summation is over the sequence of functions .
- The integration is over the closed interval .
Reasoning, step by step
- Start with the integral of the infinite sum of functions: .
- Apply the DCT to move the summation outside the integral.
- Obtain the sum of the individual integrals: .
- Conclude that the two expressions are equal under the theorem's hypotheses.
Example
The board conclusion line shows .
Common misconceptions
- Assuming that the integral of a sum always equals the sum of the integrals without verifying the DCT hypotheses.
- Confusing the interchange of limits and integrals with the interchange of derivatives and integrals.
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Related questions
The three hypotheses are: (1) a sequence of functions converges to a function on the interval ; (2) there exists a dominating function such that for all and all ; and (3) the integral of the dominating function over the interval is finite, meaning .
Conditions: The theorem is presented in a calculus-oriented form rather than a full measure-theoretic proof.; The interval is a closed, bounded real interval.
The integral is computed by substituting the Maclaurin series for , using the DCT to interchange the sum and integral, integrating each monomial term using the power rule, simplifying the factorial denominator, recognizing the resulting series as the sine series, and evaluating it at the endpoints and .
Conditions: The integration interval is .; The cosine power series is .; The DCT is applicable because the series terms are dominated by on .
The sum and integral can be interchanged because the Dominated Convergence Theorem applies. The terms of the cosine power series are bounded in magnitude by on the interval , making the constant function a valid dominating function with a finite integral over the interval.
Conditions: The integration interval is .; The series terms are .; The dominating function is .
The example is included to demonstrate that the domination hypothesis alone is insufficient if the dominating function's integral diverges. It serves as a warning that if the finite-integral condition is not met, the desired interchange of integral and sum may fail.
Conditions: The example is presented as a warning illustration on the interval .; The function dominates , but diverges.
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