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a) Let $\,f\,$ be an analytic function in the punctured disk $\,\{z\;\;;\;\;0<|z-a|<r\,\,,\,r\in\mathbb R^+\}\,$ . Prove that if the limit $\displaystyle{\lim_{z\to a}f'(z)}\,$ exists finitely, then $\,a\,$ is a removable singularity of $\,f\,$

My solution and doubt: If we develop $\,f\,$ is a Laurent series around $\,a\,$ we get $$f(z)=\frac{a_{-k}}{(z-a)^k}+\frac{a_{-k+1}}{(z-a)^{k-1}}+\ldots +\frac{a_{-1}}{z-a}+a_0+a_1(z-a)+\ldots \Longrightarrow$$ $$\Longrightarrow f'(z)=-\frac{ka_{-k}}{(z-a)^{k+1}}-\ldots -\frac{a_{-1}}{(z-a)^2}+a_1+...$$ and since $\,\displaystyle{\lim_{z\to a}f'(z)}\,$ exists finitely then it must be that $$a_{-k}=a_{-k+1}=...=a_{-1}=0$$ getting that the above series for $\,f\,$ is, in fact, a Taylor one and thus $\,f\,$ has a removable singularity at $\,a\,$ .

My doubt: is there any other "more obvious" or more elementary way to solve the above without having to resource to term-term differentiating that Laurent series?

b) Evaluate, using some complex contour, the integral $$\int_0^\infty\frac{\log x}{(1+x)^3}\,dx$$

First doubt: it is given in this exercise the hint(?) to use the function $$\frac{\log^2z}{(1+z)^3}$$Please do note the square in the logarithm! Now, is this some typo or perhaps it really helps to do it this way? After checking with WA, the original real integral equals $\,-1/2\,$ and, in fact, it is doable without need to use complex functions, and though the result is rather ugly it nevertheless is an elementary function (rational with logarithms, no hypergeometric or Li or stuff).

The real integral with the logarithm squared gives the beautiful result of $\,\pi^2/6\,$ but, again, I'm not sure whether "the hint" is a typo.

Second doubt: In either case (logarithm squared or not), what would be the best contour to choose? I though using one quarter of the circle $\,\{z\;\;;\;\;|z|=R>1\}\,$ minus one quarter of the circle $\,\{z\;\;;\;\;|z|=\epsilon\,\,,0<\epsilon<<R\}\,$, in the first quadrant both, because

$(i)\,$ to get the correct limits on the $\,x\,$-axis when passing to the limits $\,R\to\infty\,\,,\,\epsilon\to 0\,$

$(ii)\,$ To avoid the singularity $\,z=0\,$ of the logarithm (not to mention going around it and changing logarithmic branch and horrible things like this!).

Well, I'm pretty stuck here with the evaluations on the different segments of the path, besides being baffled by "the hint", and I definitely need some help here.

As before: these exercises are supposed to be for a first course in complex variable and, thus, I think they should be more or less "elementary", though this integral looks really evil.

For the time you've taken already to read this long post I already thank you, and any help, hint or ideas will be very much appreciated.

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2 Answers

up vote 2 down vote accepted

For part b) I think for this type of question you are typically supposed to use the residue theorem, so here you should look for a contour that goes around $z=-1$. Since the real integral is from $0\to \infty$ we'll likely still want to integrate along the real axis, around the circle $|z|=R$, then back along below the real axis with a small loop to avoid $z=0$. $$ \int_{x=\epsilon}^R f(x^+)+\int_{\theta= 0}^{2\pi}f(Re^{i\theta})-\int_{x=\epsilon}^R f(x^-)-\int_{\theta=0}^{2\pi}f(\epsilon e^{i\theta}) = 2\pi i~\mathrm{Res}(f,-1) $$ Using the hint and taking $R\to\infty,\epsilon\to 0$ the circular integrals vanish, the first integral along the real line has $\log^2x$ in the numerator, and the second has the next branch value $(\log x+2\pi i)^2$. When you take the difference the $\log^2 x$ terms cancel and you end up with your desired integral in terms of another integral that is easy to evaluate and the residue.

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Thank you for the great answer,@Zander. Something like this I thought of after posting the question, but it looked too malicious for an introductory course...go figure! +1 –  DonAntonio Jul 1 '12 at 9:58
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For a). Since $f'$ has a removable singularity at $a$, there is an analytic function $g$ on $\{z:|z-a|<r\}$ such that $f'=g$ on the punctured disk. Analytic functions on disks have antiderivatives; this can be seen by integrating the power series term by term, or by showing that $G(z)=\int_{[a,z]}g(z)dz$ is an antiderivative of $g$. (Here $[a,z]$ is used to denote the directed line segment from $a$ to $z$.) So there is an analytic function $G$ on the disk with $G'=g=f'$ on the punctured disk. This implies that $f-G$ is equal to a constant $c$, hence $\lim\limits_{z\to a}f(z)=c+G(a)$ exists.

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Nice answer, @Jonas. Thank you. +1 –  DonAntonio Jul 1 '12 at 9:38
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