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I would like to know if there is a way how to calculate the following integrals: $$ \int_{0}^{\pi/2} \ln(1-\cos x) \mathrm dx $$ as well as $$ \int_{0}^{\pi/2} \ln(1-\sin x) \mathrm dx $$ Actually both integrals are equal to each other (which can be shown through that integral property $x→\pi/2-x$ substitution, sorry got no formal name for it). And so their difference results in zero. But is there a tricky way to evaluate one of them? According to my TI it is around -2.920558 and so I suspect they are not divergent, as I first thought. Thanks in advance

Thomas Andrews
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imranfat
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  • It's not very easy to integrate... – Shahar Mar 10 '14 at 21:40
  • Symbolic calculators tell me that the answer is $-2 C-\frac{1}{2} \pi \log2$, where $C$ is Catalan's constant. That makes me suspect that the way to go is to expand $\log(1-u)$ into its power series, plug in $u=\cos x$, and try integrating term by term.... – Greg Martin Mar 10 '14 at 22:06
  • @GregMartin I wasn't really expecting a fine anti derivative. I was hoping, that since this is a definite integral, that these limits might be used to do the trick. After all, through the cofunction identity sine and cosines can be written in terms of one another. I was hoping that something like that could be used. The series may indeed be something else I could explore (tomorrow...) – imranfat Mar 10 '14 at 22:13
  • Or perhaps use $1-\cos x=2\sin^2 x/2$ and $\sin z=z\prod_{k\in\Bbb Z\setminus{0}}(1-\tfrac{z}{k\pi})$. – Lutz Lehmann Mar 10 '14 at 23:46
  • @GregMartin: I tried that, but you get a complicated series involving the Beta function. The approach I finally took was to write $1-\cos(x)=\dfrac{(1-e^{ix})(1-e^{-ix})}{2}$ then use the series for $\log(1+x)$. – robjohn Mar 11 '14 at 14:16

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$$ \begin{align} \int_0^{\pi/2}\log(1-\cos(x))\,\mathrm{d}x &=\int_0^{\pi/2}\left[\log(1-e^{ix})+\log(1-e^{-ix})-\log(2)\right]\,\mathrm{d}x\\ &=-\frac\pi2\log(2)-2\int_0^{\pi/2}\sum_{k=1}^\infty\frac{\cos(kx)}{k}\,\mathrm{d}x\\ &=-\frac\pi2\log(2)-2\sum_{k=1}^\infty\frac{\sin(k\pi/2)}{k^2}\\ &=-\frac\pi2\log(2)-2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\\ &=-\frac\pi2\log(2)-2\mathrm{G} \end{align} $$ where $\mathrm{G}$ is Catalan's Constant.

robjohn
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  • Thanks for the effort. It just doesn't come out nice unfortunately....Can't use Catalan's in my class :) – imranfat Mar 11 '14 at 14:25
  • It's either use Catalan's constant or the series for it. I don't think you'll find anything simpler. – robjohn Mar 11 '14 at 14:28
  • Ah, I see. Your method gives integrals of $\cos kx$ rather than the $\cos^kx$ straight from the power series. The latter wouldn't be insurmountable, but this is more direct, I agree. Well done! – Greg Martin Mar 11 '14 at 17:03