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I would like to know if it is true that if one term (summand) diverges then the whole improper integral diverges. I would have thought that it would be like splitting a fraction, that divergance of the numerator can/may be compensated by the denominator, so it tells you nothing, that i would have to check if the other is finite.

If i split an improper integral:

$$\int_a^b fdx = \int_a^b g+h dx = \int_a^b g dx + \int_a^bh dx$$

then the Prof. says that if i can show that $\int_a^b gdx$ diverges then i dont even have to check the other summand.

(I realize that the statement would be true if we were considering absolute values but the Prof. did not mention absolute value/convergence)

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Let $\int_a^b f\,dx$ converge, and $\int_a^b b\,dx$ diverge. Then what about $g = (f - b)$, $h = b$? – Daniel Fischer Sep 28 '13 at 12:06
@DanielFischer Let $\int b$ also diverge but to the opposite infinity, does +inf-inf not have a chance to converge? – ben Sep 28 '13 at 12:10
Well, that's my point, the badnesses can cancel. They do in my example. – Daniel Fischer Sep 28 '13 at 12:12
up vote 3 down vote accepted

It is true under extra conditions. For instance that the summands are non-negative. If it would be true without conditions then every integral would diverge.

Let it be that $\int fdx$ diverges. Then $\int gdx=\int\left(g-f\right)+fdx$ would diverge too.

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Thats what i thought. Thanks. – ben Sep 28 '13 at 12:28

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