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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.

Arturo Magidin
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mary
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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? – Qiaochu Yuan Apr 24 '11 at 00:45

1 Answers1

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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.

Ross Millikan
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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? – mary Apr 24 '11 at 01:03
  • I am not understanding the second part – mary Apr 24 '11 at 01:03
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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. – Ross Millikan Apr 24 '11 at 01:34
  • Second hint: If you want to make a function that integrates this way, how about a series of steps? – Ross Millikan Apr 24 '11 at 02:26
  • How can we transform f to get g? What sort of transformation are we looking for? – mary Apr 24 '11 at 04:56
  • 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) – Ross Millikan Apr 24 '11 at 05:12
  • f = log(2); I am still not getting the next part, can you explain? – mary Apr 24 '11 at 05:40
  • @user8917 $\int_1^\infty \log 2 dx$ is unbounded! – Glen Wheeler Apr 24 '11 at 08:41
  • Maybe we want to mollify this a little bit by replacing the alternating sign by a $\sin$-wave and the floor-function by $x\ $? – Christian Blatter Apr 24 '11 at 09:36
  • So how about f(x)=$\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? – Ross Millikan Apr 24 '11 at 15:10
  • 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 section – mary Apr 24 '11 at 22:24
  • 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? – mary Apr 25 '11 at 04:14
  • Also, I am still not getting how to transform f to g – mary Apr 25 '11 at 04:15
  • 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. – Ross Millikan Apr 25 '11 at 04:26
  • Can you please put this in the answer section, it is a bit hard to read this in the comments thanks – mary Apr 25 '11 at 04:29
  • 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. – Ross Millikan Apr 25 '11 at 04:37
  • 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? – mary Apr 25 '11 at 04:52
  • 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. – Ross Millikan Apr 25 '11 at 12:50
  • 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 about – mary Apr 26 '11 at 06:44