16

$$\int_0^{\pi/2} \ln{(x^2 + \ln^2{(\cos{x})})} \,\mathrm{d}x$$

I was given this integral yesterday by someone on a forum and after a few hours of having a go at it I didn't really get anywhere significant.

My first idea was to use a series of substitutions which would simply the integral into a form where taylor expansions could be used to solve it. Another idea of mine was to use complex numbers to get rid of some of the log terms, changing the $\ln^2{(\cos{x})}$ term could possibly make this integral a lot more manageable.

I'd prefer if someone could help me solve this using basic methods although if there is a more complicated but elegant solution (using contour integration for example) it could still be beneficial to someone else. We don't do the equivalent of Calc III in my school so I am not very familiar with methods which go beyond the Calc II syllabus (DUDIS, Laplace ect..).

GNUSupporter 8964民主女神 地下教會
  • 17,627
Loua
  • 347
  • 3
    Why do you believe that there is a closed-form solution? – Mark Viola Jan 30 '16 at 16:14
  • 1
    @Dr.MV It was given to me after I had spent quite a while on it that the closed form was $\pi \ln{(\ln{2})}$, I was also told that it was possible using methods I was familiar with. – Loua Jan 30 '16 at 16:17
  • 3
    I think the one idea here would be to use something like $x^2+\log(\cos(x))=|\log(\frac{1+e^{2 i x}}{2})|^2$ and map everything ($z=e^{ix}$) to the unit circle (or some deformed version of it) to employ residue theorem. maybe someone else may finish it from here, i'm much to busy today ... – tired Jan 30 '16 at 16:30
  • 2
    Actually the $2\pi i \times \text{Res}(z=0)=\pi \log(\log(2))$ so this seems definitly a way to go! Have fun with it guys :) – tired Jan 30 '16 at 16:34
  • 7
    Other evaluations are here: http://math.stackexchange.com/questions/915175/prove-displaystyle-int-0-pi-2-ln-leftx2-ln-cos-x2-right – Noam Shalev - nospoon Jan 30 '16 at 17:20
  • @tired Nobody is deleting those solutions, or removing their upvotes. –  Jan 30 '16 at 22:26
  • @LiveForever u are totally right... writing-> thinking is never a good idea ^^ – tired Jan 30 '16 at 22:43
  • I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 17:08

2 Answers2

10

Note that the integral is

$$2 \operatorname{Re}{\int_0^{\pi/2} dx \, \log{(\log{\cos{x}} + i x)}} $$

which, by using various manipulations of the integrand around the unit circle, we find to be

$$\frac12 \operatorname{Re}{\int_{-\pi}^{\pi} dx \, \log{(\log{\cos{x}} + i x)}} $$

which integral we may express as an integral over a slightly deformed unit circle:

$$-\frac{i}{2} \oint_C \frac{dz}{z} \log{\left [\log{\left(\frac{z+z^{-1}}{2} \right)}+\log{z} \right ]} = -\frac{i}{2} \oint_C \frac{dz}{z} \log{\left [\log{\left(\frac{z^2+1}{2} \right)} \right ]}$$

where $C$ is the unit circle with small semicircular indentations at the poles $z=\pm i$. The contributions from these indentations to the integral is zero as the radius of the indentations goes to zero.

Thus, we have the integral of a holomorphic function over the boundary of a region in which there is a single, simple pole at the origin. By Cauchy's theorem/ the residue theorem, the integral we seek is thus the real part of

$$i 2 \pi \left (-\frac{i}{2} \right ) \log{\log{\frac12}} $$

or $\pi \log{\log{2}} $.

ADDENDUM

Let me fill in the gaps. Imagine

$$I = \int_{0}^{\pi/2} dx \, \log{(\log{\cos{x}} + i x)}$$

and

$$\begin{align}I' &= \int_{\pi/2}^{\pi} dx \, \log{(\log{\cos{x}} + i x)} \\ &= \int_{0}^{\pi/2} dx \, \log{\left [\log{\left (-\cos{\left (\frac{\pi}{2} - x \right )}\right )} + i x+i \frac{\pi}{2}\right ]} \\ &= \int_{0}^{\pi/2} dx \, \log{\left [\log{\left (\cos{x}\right )} - i \pi + i \left (\frac{\pi}{2}-x \right )+i \frac{\pi}{2}\right ]}\\ &= \int_{0}^{\pi/2} dx \, \log{(\log{\cos{x}} - i x)}\end{align}$$

Then

$$I+I' =\int_0^{\pi} dx \, \log{(\log{\cos{x}} + i x)}= \int_0^{\pi/2} dx \, \log{(\log^2{\cos{x}}+x^2)} $$

Thus,

$$\operatorname{Re}{\int_{-\pi}^{\pi} dx \, \log{(\log{\cos{x}} + i x)}} = 2 \int_0^{\pi/2} dx \, \log{(\log^2{\cos{x}}+x^2)}$$

as was to be shown.

Ron Gordon
  • 138,521
4

Following the comment left by @tired, we note that we can factor the argument of the quadratic as

$$x^2+\log^2(\cos(x))=(\log(\cos(x)+ix)(\log(\cos(x)-ix)$$

Writing $ix=\log(e^{ix})$ we have

$$x^2+\log^2(\cos(x))=\left|\log(\cos(x)e^{ix})\right|^2=\left|\log\left(\frac{1+e^{i2x}}{2}\right)\right|^2$$

Next, we write the integral of interest as

$$\int_0^{\pi/2}\log(x^2+\log^2(\cos(x)))\,dx=\frac14\int_{-\pi}^{\pi}\log\left(\left|\log\left(\frac{1+e^{ix}}{2}\right)\right|^2\right)\,dx$$

We move to the complex plane by letting $z=e^{ix}$. The integral of interest becomes

$$\begin{align} \int_0^{\pi/2}\log(x^2+\log^2(\cos(x)))\,dx&=\frac12\oint_{|z|=1}\log\left(\left|\log\left(\frac{1+z}{2}\right)\right|\right)\frac{1}{iz}\,dz\\\\ &=\frac12\text{Re}\left(\oint_{|z|=1}\log\left(\log\left(\frac{1+z}{2}\right)\right)\frac{1}{iz}\,dz\right)\\\\ &=\frac12 \text{Re}\left(2\pi i \left(\frac{\log(\log(2))+i\pi}{i}\right)\right)\\\\ &=\pi \log(\log(2)) \end{align}$$

where we tacitly deformed the contour around the branch point singularities at $z=1$ and $z=-1$ and noted that the contributions from integrations around the deformations are zero.

Mark Viola
  • 179,405