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Everybody Hello, I'm confused about the following integral:

Consider the following:
Riemann Integral: $\int f dx:=\{\sum_{E\in\mathcal{E}} f(E)\lambda(E)\}_\mathcal{E}$
Domain: $I:=[0,1]$
Function: $f(x):=\frac{1}{\sqrt x}$, $f(0):=0$

The Problem is now:
This should be Riemann integrable on the compact unit interval as we know by formal antiderivative and neglecting the boundary, however, comparing upper and lower sum or equivalently tagged partitions it shouldn't be Riemann integrable ...did I miss sth?

That is in Detail:
Take the refining sequence of partitions: $\mathcal{E}_n:=\{(\frac{k-1}{n},\frac{k}{n}]\}_{k=1}^n$
Then compute the upper sum by taking the suprema: $\sum_{k=1}^n\sup f(E_k)\lambda(E_k)=\infty\frac{1}{n}+\sqrt{\frac{n}{1}}\frac{1}{n}+\sqrt{\frac{n}{2}}\frac{1}{n}\ldots=\infty$
...this already tells us it shouldn't be Riemann integrable

Moreover, It seems as if the function even fails to be integrable as improper integral.
Am I missing sth when simply considering equidistant partitions? Doing so I end up with sum variaty of the Riemann Zeta Function being divergent...

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  • $\begingroup$ Put here your calculations for "... comparing upper and lower sum..." $\endgroup$ – Martín-Blas Pérez Pinilla Feb 8 '14 at 14:32
  • $\begingroup$ Since $f$ is not bounded it is not Riemann integrable on $[0,1]$. The known finite value of the intended integral results from considering it as an improper integral resulting from integrals over $[\epsilon,1]$ and passing to the limit $\epsilon\to 0+$. $\endgroup$ – Christian Blatter Feb 8 '14 at 14:43
  • $\begingroup$ Ah right, thx! ...btw, are you THE prof. christian blatter??? small fan of you =D $\endgroup$ – C-Star-W-Star Feb 8 '14 at 14:54
  • $\begingroup$ It seems as if the function isn't even integrable for $\epsilon>0$. Am I missing sth when simply considering equidistant partitions? $\endgroup$ – C-Star-W-Star Feb 26 '14 at 15:41
  • $\begingroup$ For $\epsilon > 0$, the function is Riemann-integrable on $[\epsilon,1]$. What problem do you see when trying to prove that? $\endgroup$ – Daniel Fischer Feb 26 '14 at 16:01
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Ok, so the function is improperly riemann integrable. While considering partitions one has to be careful with taking limit same time

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