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The integral is .
is discontinuous at .
If is continuous on the interval , except for some in at which has an infinite discontinuity, then .
The improper integral on the left diverges if either of the improper integrals on the right diverges.
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Apply formula: .
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Therefore, the series is converges to .
Graph:
Graph the integrand .
Observe the graph:
Integration capabilities: .
The improper integral is converges to .
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