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In a proof of the Basel problem, the excellent YouTuber "blackpenredpen" relies on a manipulation which I assume is valid, but I don't know why it is valid.

They split the following integral: $$\int_0^{\pi/2}\ln(2\cos x)\space dx=\int_0^{\pi/2}\ln(e^{ix}+e^{-ix})\space dx=$$ $$\int_0^{\pi/2}\ln(e^{ix}(1+e^{-2ix}))\space dx=\int_0^{\pi/2}\ln(e^{ix})\space dx + \int_0^{\pi/2}\ln(1+e^{-2ix}))\space dx$$

And in the calculation of the second part of the integral on the right hand side, they use the series expansion:

$$\ln(1+x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}{n}$$

to represent $\ln(1+e^{-2ix})$ and note that this is only valid for $|x|\leq1, x\neq-1$. Although $|e^{-2ix}|=1,\forall x$, I noticed that the integral runs up to $\frac{\pi}{2}$, and $e^{-2i\cdot\pi/2}=-1$, which means that the series expansion does not converge (the logarithm goes to $\ln(0)$ which is very undefined!) at the upper bound of the integral. I imagine this is valid due to integrals being limits, and perhaps we "approach" $\frac{\pi}{2}$ without reaching it, but I'd like a formal explanation for why this expansion is valid here - I would like to learn when we can and cannot do this sort of thing!

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  • $\begingroup$ The remainder term in the Taylor series is relevant here. The Taylor series to $n$ terms involves the first $n$ derivatives, then the remainder term involves the $n+1$th. Then you try to show the remainder approaches zero. $\endgroup$
    – Empy2
    Jun 6 at 12:55
  • $\begingroup$ The original integral is improper, since the integrand has a singularity at $x=\pi/2$. So the integral is defined to be $\lim_{t\nearrow \pi/2}\int_0^t \ln(2\cos x)\,dx$. (I'm assuming these are Riemann integrals, improper integrals are treated differently in e.g. Lebesgue integration) Separately, it might be a little risky to assume a manipulation is valid just because it arrives at the correct answer. $\endgroup$ Jun 7 at 20:05
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    $\begingroup$ this logic is flawed: "relies on a manipulation which is clearly valid, since their proof produces the correct answer" $\endgroup$ Jul 7 at 19:09
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    $\begingroup$ Arriving at the right answer does not mean that one's technique is valid. For example, in $16/64,$ if one cancels the $6$ from the numerator and the denominator, one gets $1/4.$ And it is correct that $16/64$ reduced to lowest terms is $1/4. \qquad$ $\endgroup$ Jul 7 at 19:19
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    $\begingroup$ @jimjim Wolfram Alpha would seem to disagree with you and labels it as undefined. The polar form argument hasn’t led me anywhere: $0^i=(re^{i\theta})^i=(r^i)(e^{-\theta})=0^i\cdot e^{-\theta}$ where $\theta$ could be anything and we are still left with $0^i$. $\endgroup$
    – FShrike
    Jul 13 at 8:47
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For $\epsilon>0$, we wish to evaluate $\int_0^{\pi/2-\epsilon}\ln(1+e^{-2ix})dx$. Let $\delta>0$ be small, so that

\begin{align} \int_0^{\pi/2-\epsilon}\ln(1+e^{-2ix})dx=\int_0^{\pi/2-\epsilon}\ln(1+(1-\delta)e^{-2ix})dx+\int_0^{\pi/2-\epsilon}\ln\left(\frac{1+e^{-2ix}}{1+(1-\delta)e^{-2ix}}\right)dx. \end{align} Here, \begin{align} \left|\ln\left(\frac{1+e^{-2ix}}{1+(1-\delta)e^{-2ix}}\right)\right|&=\left|\ln\left(1-\frac\delta{1+e^{2ix}}\right)\right|\\ &\le\frac32\frac\delta{|1+e^{2ix}|}\\ &=\frac32\frac\delta{\sqrt{(1+\cos(2x))^2+\sin^2(2x)}}\\ &=\frac32\frac\delta{\sqrt{2+2\cos(2x)}},\\ \end{align} (here, I used this bound), so for $\delta$ small enough, the second term disappears, since $2+2\cos x\ge 2-2\cos(2\epsilon)>0$. Now, one can safely trade the limit and the integral because of uniform convergence:

\begin{align} \int_0^{\pi/2-\epsilon}\ln(1+(1-\delta)e^{-2ix})dx&=\int_0^{\pi/2-\epsilon}\lim_{N\to\infty}\sum_{n=1}^N(-1)^{n-1}\frac1n(1-\delta)^ne^{-2nix}dx\\ &=\lim_{N\to\infty}\sum_{n=1}^N(-1)^{n-1}\frac1n(1-\delta)^n\int_0^{\pi/2-\epsilon}e^{-2nix}dx\\ &=\sum_{n=1}^\infty(-1)^{n-1}\frac1{2n^2i}(1-\delta)^n(1-(-1)^ne^{2ni\epsilon}). \end{align} Since $\epsilon,\delta>0$ were arbitrary, we can take the limit as they approach $0$.

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  • $\begingroup$ Very thorough thank you,... is the takeaway for me that in any improper integral, I must manually check if the limits exist as we closer to the improper bound? There is no intuitive or good rule as such to determine the validity of such an integral? $\endgroup$
    – FShrike
    Jul 12 at 10:58
  • $\begingroup$ Also, your placement of $\delta$ in the original integral didn’t affect the integral at all, so I’m not sure what purpose $\delta$ has; surely one could just take the limit as $\epsilon$ approaches zero... as delta varies the integral expression never varied, is what I mean to say $\endgroup$
    – FShrike
    Jul 12 at 11:04
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    $\begingroup$ The point of $\delta$ was to be able to exchange the limit and the integral. Uniform convergence requires the terms in the series to decrease quickly, and $\frac1ne^{-2nix}$ doesn't (but $\frac1n(1-\delta)^ne^{-2nix}$ does). $\endgroup$
    – Kenta S
    Jul 12 at 12:00
  • $\begingroup$ Do you happen to know the name or origin of the statement you linked on the Wikipedia page? It seems a very useful result, and I'd like to see a proof (but as usual Wikipedia just states the fact without showing where it came from) $\endgroup$
    – FShrike
    Jul 12 at 13:31
  • $\begingroup$ @IdioticShrike it is an easy exercise in $\epsilon-\delta$. $\endgroup$
    – Kenta S
    Jul 12 at 15:00

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