Logarithms, Trignometry, Integrals (Part 3)

Firstly, the substitution $u = \tan x$ gives us $\int_0^{\frac12\pi} \ln\big(\big\vert\cos(\tan x)\big\vert \big)\,dx \; = \; \int_0^\infty \frac{\ln \vert \cos u\vert}{u^2+1}\,du \; = \; \tfrac12\int_0^\infty \frac{\ln(\cos^2u)}{u^2+1}\,du$ We could look this last integral in tables of integrals, but let's work it out. If $\mathrm{Im}z > 0$ then, since $|e^{2iz}| = e^{-2\mathrm{Im}z} < 1$ , $1 + e^{2iz}$ has positive real part. Thus we can define $\log(\tfrac12(1+e^{2iz}))$ for any complex number with positive imaginary part, using the principal branch of the logarithm. In addition $\log(\tfrac12(1 + e^{2iz})) + \log(\tfrac12(1 + e^{2iw})) \; = \; \log\big(\tfrac14(1 + e^{2iz})(1 + e^{2iw})\big)$ for any $z,w$ with positive imaginary part. The function $F(z) \; = \; \frac{\log(\frac12(1 + e^{2iz}))}{z^2+1}$ is analytic for the upper half plane of complex numbers with positive imaginary part, with the point $i$ removed.

For any $0 < \varepsilon < 1$ and integer $N \ge 1$ , let $\Gamma_{\epsilon,N}$ be the contour $\gamma_1 + \gamma_R + \gamma_2 - \delta_\epsilon$ consisting of

• the line segment $\gamma_1 = [\epsilon,\pi N]$ ,
• the anticlockwise semicircular arc $\gamma_N$ given by $z \,=\, \pi N e^{i\theta}$ for $0 \le \theta \le \pi$ ,
• the line segment $\gamma_2 = [-\pi N,-\epsilon]$ ,
• the anticlockwise semicircular arc $\delta_\epsilon$ given by $z \,=\, \epsilon e^{i\theta}$ for $0 \le \theta < \pi$ .

Then $\int_{\Gamma_{\epsilon,N}}F(z)\,dz \; = \; 2\pi i \mathrm{Res}_{z=i} F(z) \; = \; 2\pi i \dfrac{1}{2i}\ln\big(\tfrac12(1 + e^{-2})\big) \; = \; \pi \ln\big(\tfrac12(1 + e^{-2})\big)$ Now $\begin{array}{rcl} \displaystyle\left(\int_{\gamma_1} + \int_{\gamma_2}\right)F(z)\,dz & = & \displaystyle \left(\int_{\epsilon}^{\pi N} + \int_{-\pi N}^{-\epsilon}\right) F(x)\,dx \; = \; \int_\epsilon^{\pi N} \big(F(x) + F(-x)\big)\,dx \\ & = & \displaystyle \int_\epsilon^{\pi N} \frac{\log\big(\frac12(1 + e^{2ix})\big) + \log \big(\frac12(1 + e^{-2ix})\big)}{x^2+1}\,dx \\ & = & \displaystyle\int_\epsilon^{\pi N} \frac{\log\big(\tfrac14(1 + e^{2ix})(1 + e^{-2ix})\big)}{x^2+1}\,dx \; = \; \int_\epsilon^{\pi N} \frac{\ln\big(\tfrac12(1 + \cos2x)\big)}{x^2+1}\,dx \\ & = & \displaystyle\int_\epsilon^{\pi N} \frac{\ln(\cos^2x)}{x^2+1}\,dx \end{array}$ Now the periodicity property of $e^{2iz}$ can be used to show that there exists $k > 0$ such that $k \le \big\vert\tfrac12(1 + e^{2iz})\big| \le 1$ for all $z \in \gamma_N$ for all $N \in \mathbb{N}$ , which implies that $\log\big(\tfrac12(1 + e^{2iz})\big)$ is uniformly bounded on $\gamma_N$ for all $N$ . This implies that $\int_{\gamma_N} F(z)\,dz \; = \; O\big(N^{-1}\big) \qquad \qquad N \to \infty \;.$ On the other hand, it is possible (if a bit fiddly) to show that $\int_{\delta_\epsilon} F(z)\,dz \; = \; O\big(\epsilon \ln \epsilon\big) \qquad \qquad \epsilon \to 0 \;.$ Letting $\epsilon \to 0$ and $N \to \infty$ , we deduce that $\int_0^{\frac12\pi} \ln\big(\big\vert\cos(\tan x)\big\vert \big)\,dx \; = \; \tfrac12\int_0^\infty \frac{\ln(\cos^2u)}{u^2+1}\,du \; = \; \tfrac12\pi\ln\big(\tfrac12(1 + e^{-2})\big)$ making the answer $1 + 2 + 1 + 2 + 2 = 8$ .

Put $\tan{x}=y$ , to get our integral as :

$\displaystyle I = \frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ \frac { \ln { \cos ^{ 2 }{ x } } }{ 1+{ x }^{ 2 } } dx }$

Now using the fourier expansion of $\displaystyle \frac { 1 }{ 2 } \ln { \cos ^{ 2 }{ x } } =\sum _{ k=1 }^{ \infty }{ \frac { { (-1) }^{ k-1 } }{ k } \cos { (2kx) } - \ln(2) }$ , we get :

$\displaystyle I = \int _{ 0 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \frac { { (-1) }^{ k-1 } }{ k } \frac { \cos { (2kx) } }{ 1+{ x }^{ 2 } } } dx } - \dfrac{\pi \ln(2)}{2}$

Interchanging summation and integral we have :

$\displaystyle I = \sum _{ k=1 }^{ \infty }{ \frac { { (-1) }^{ k-1 } }{ k } \int _{ 0 }^{ \infty }{ \frac { \cos { (2kx) } }{ 1+{ x }^{ 2 } } dx } }- \dfrac{\pi \ln(2)}{2}$

Using the well known result $\displaystyle \int _{ 0 }^{ \infty }{ \frac { \cos { (2kx) } }{ 1+{ x }^{ 2 } } dx } =\frac { \pi }{ 2{ e }^{ 2k } }$

We get our integral as :

$\displaystyle I = \sum _{ k=1 }^{ \infty }{ \frac { { (-1) }^{ k-1 } }{ k } \frac { \pi }{ 2{ e }^{ 2k } } } - \dfrac{\pi \ln(2)}{2}$

$\displaystyle I = \frac { \pi }{ 2 } \sum _{ k=1 }^{ \infty }{ \frac { { (-1) }^{ k-1 }{ e }^{ -2k } }{ k } } -\frac { \pi }{ 2 } \ln { (2) } =\frac { \pi }{ 2 } \ln { (1+{ e }^{ -2 }) } -\frac { \pi }{ 2 } \ln { (2) }$

$\Rightarrow \boxed{\displaystyle I =\frac { \pi }{ 2 } \ln { \left( \frac { 1+{ e }^{ -2 } }{ 2 } \right) } }$