If a partition of an integral diverges, does the whole integral diverge?

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The discussion centers on the divergence of integrals, specifically addressing the question of whether the divergence of a partitioned integral implies the divergence of the entire integral. It is established that if one of the integrals in the partition diverges and the partition point is between the limits of integration, the overall integral must also diverge. However, if the partition point lies outside the limits, the overall integral may still converge, as demonstrated with the example of \(\int_{1}^{2} \frac{1}{x}dx\) converging while \(\int_{1}^{-1}\frac{1}{x}dx + \int_{-1}^{2} \frac{1}{x}dx\) diverges due to the presence of an unbounded region.

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\int^{b}_{a}f(x)dx = \int^{c}_{a}f(x)dx + \int^{b}_{c}f(x)dx

If one of the integrals on the right-hand-side is known to diverge, must the integral on the left also necessarily diverge?

BiP
 
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Yes, by definition of improper integrals.
 
Yes if c is between a and b. But not necessarily if c is not between a and b:

\int_{1}^{2} \frac{1}{x}dx converges, but \int_{1}^{-1}\frac{1}{x}dx + \int_{-1}^{2} \frac{1}{x}dx diverges (each term is an integral over a region containing 0, where 1/x is unbounded).
 

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