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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 "find an example and prove that it works." What do you know about the conditional convergence of alternating series?2011-04-24

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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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    So, for the first part, if I consider log(2), then I would need to prove this and somewhat need to use Abel's Theorem?2011-04-24
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    I am not understanding the second part2011-04-24
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    @user8917: no, can you find a function that integrates to make the series for log(2)? The negative terms come because the function is less than zero. Then when you take the absolute value of $f$, the negative terms become positive and the integral diverges like the harmonic series.2011-04-24
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    Second hint: If you want to make a function that integrates this way, how about a series of steps?2011-04-24
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    How can we transform f to get g? What sort of transformation are we looking for?2011-04-24
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    Does that mean you have an $f$? What is it? Not necessarily a transformation, as the same way of thinking about the problem. You don't have as much room (only one unit instead of $\infty$, so each step has to get its area more quickly)2011-04-24
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    f = log(2); I am still not getting the next part, can you explain?2011-04-24
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    @user8917 $\int_1^\infty \log 2 dx$ is unbounded!2011-04-24
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    Maybe we want to mollify this a little bit by replacing the alternating sign by a $\sin$-wave and the floor-function by $x\ $?2011-04-24
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    So how about f(x)=$\frac(-1)^{lfloor x \rfloor}{lfloor x \rfloor}$? Each segment is one unit long and they alternate in sign. But if you take the absolute value, you get the harmonic series. Now how can you pack the same into a single unit of length?2011-04-24
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    I am still not understanding why you use floor of x. Can you post it on the answer thread, it is a bit hard to read this in the comments section2011-04-24
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    So to finish part a, I could try something like what you have for f, using floors and ceilings. But how to prove that it works?2011-04-25
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    Also, I am still not getting how to transform f to g2011-04-25
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    If you split $\int_1^{\infty}=\int_1^2+\int_2^3+\ldots$, you should be able to do each piece and see it works. Now for $f$ you had a bunch of $1 \times $\frac{1}{n}$ rectangles to integrate. You can't just rotate 90 degrees, as the sum of the harmonic series is greater than 1, but maybe you can find some other rectangles that fit within (0,1] in $x$ that have the same areas.2011-04-25
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    Can you please put this in the answer section, it is a bit hard to read this in the comments thanks2011-04-25
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    If you split $\int_1^{\infty}=\int_1^2+\int_2^3+\ldots$, you should be able to do each piece and see it works. Now for f you had a bunch of $1×\frac{1}{n}$ rectangles to integrate. You can′t just rotate 90 degrees,as the sum of the harmonic series is greater than 1,but maybe you can find some other rectangles that fit within(0,1] in x$ that have the same areas. Have you had any thoughts? I haven't seen any evidence.2011-04-25
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    I am unsure about the rotation and why we are considering to use f to get g. Can we be more inovative to get g?2011-04-25
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    You're not rotating it, you are just using the same technique of finding appropriate rectangles to make the integral come out like you want.2011-04-25
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    Can 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