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Prove there exists a function $f$ such that $\int_1^{\infty}f(x)\,dx\text{ converges, but }\int_1^{\infty}|f(x)|\,dx\text{ diverges.}$

Similarly, prove that there exists a function $g$ such that $\int_0^1 g(x)\,dx\text{ converges, but }\int_0^1|g(x)|\,dx\text{ diverges.}$

All I am able to understand in the first part, is to take an example. I am thinking of something like $(1/2)^n$? I am not sure how to account for the absolute values, and when they say prove, can I just find an example only? I am having trouble of thinking of such a function.

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    "Prove" in this context means "$f$ind an example and prove that it works." What do you know about the conditional convergence of alternating series?2011-04-24

1 Answers 1

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Hint: for the first part, think of the series $\log 2=1-\frac{1}{2}+\frac{1}{3}-\ldots$. It converges (and we know to what) but taking the absolute value of each term yields the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\ldots$ which diverges. Can you turn that into an integral? Yes, one way to prove something exists is to exhibit it. For the second part, informally $0=\frac{1}{\infty}$, so maybe you can transform your $f$ in some way to get $g$.

Added: Try $f(x)=\frac{(-1)^{\lfloor x+1 \rfloor}}{\lfloor x \rfloor}$ (I missed some formatting in the comment). If you integrate this from $1$ to $\infty$, each segment of the form $[n,n+1)$ gives one term in the expansion of $\log 2$. Then taking the absolute value gives the harmonic series.

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    Ca$n$ you explain how to get g, and please put this in the answer section, the text is hard to read. I am also not following the same technique you are talking about2011-04-26