Definte Integral #2

[sbhatnagar]

Prove that

\[\int_0^{\pi/2}\cos(\ln(\tan(x)))dx=\frac{\pi}{2} \text{sech} \left(\frac{\pi}{2}\right)\]

 

[Bacterius]

Spoiler

Let $u = \ln{\tan{x}}$. Then we have $x = \arctan{(e^u)}$ and:

$$\mathrm{d} x = \frac{1}{e^u + e^{-u}} ~ \mathrm{d} u = \frac{1}{2} \mathrm{sech}{(u)} ~ \mathrm{d} u$$

We also have the new integral bounds $\ln{(\tan{0})} = -\infty$ and $\ln{(\tan{\frac{\pi}{2}})} = + \infty$. Substituting:

$$\displaystyle \int_0^{\frac{\pi}{2}} \cos{\ln{\tan{x}}} ~ \mathrm{d} x = \frac{1}{2} \int_{-\infty}^{+\infty} \cos{(u)} \mathrm{sech}{(u)} ~ \mathrm{d} u$$

We note the integrand is even, as both $\cos{(u)}$ and $\mathrm{sech}{(u)}$ are even. Thus we have:

$$\frac{1}{2} \int_{-\infty}^{+\infty} \cos{(u)} \mathrm{sech}{(u)} ~ \mathrm{d} u = \int_{0}^{+\infty} \cos{(u)} \mathrm{sech}{(u)} ~ \mathrm{d} u$$

Now substitute $z =e^u$, so $u = \ln{z}$, and hence $\mathrm{d} u = \frac{1}{z} ~ \mathrm{d} z$. The bounds now become $e^0 = 1$ and $e^{+\infty} = + \infty$. Substituting:

$$\int_{0}^{+\infty} \cos{(u)} \mathrm{sech}{(u)} ~ \mathrm{d} u = \int_{1}^{+\infty} \cos{(\ln{z})} \mathrm{sech}{(\ln{z})} \cdot \frac{1}{z} ~ \mathrm{d} z$$

Using the complex exponential form of the hyperbolic secant, we see that:

$$\mathrm{sech}{(\ln{z})} = \frac{2z}{z^2 + 1}$$

And thus, our integral becomes:

$$\int_{1}^{+\infty} \frac{2z \cos{\ln{z}} }{z^2 + 1} \cdot \frac{1}{z} ~ \mathrm{d} z = 2 \int_{1}^{+\infty} \frac{\cos{\ln{z}} }{z^2 + 1} ~ \mathrm{d} z$$

We know from the complex exponential form of $\cos$ that:

$$\cos{\ln{z}} = \frac{1}{2} \left ( z^{-i} + z^i \right )$$

Substituting this into the integrand, we get:

$$\int_{1}^{+\infty} \frac{z^{-i} + z^i}{z^2 + 1} ~ \mathrm{d} z$$

And I can't seem to get any farther. I'll try and keep going later, I need sleep. Am I even on the right track, anyway?

 

[sbhatnagar]

Spoiler

Let $u = \ln{\tan{x}}$. Then we have $x = \arctan{(e^u)}$ and:

$$\mathrm{d} x = \frac{1}{e^u + e^{-u}} ~ \mathrm{d} u = \frac{1}{2} \mathrm{sech}{(u)} ~ \mathrm{d} u$$

We also have the new integral bounds $\ln{(\tan{0})} = -\infty$ and $\ln{(\tan{\frac{\pi}{2}})} = + \infty$. Substituting:

$$\displaystyle \int_0^{\frac{\pi}{2}} \cos{\ln{\tan{x}}} ~ \mathrm{d} x = \frac{1}{2} \int_{-\infty}^{+\infty} \cos{(u)} \mathrm{sech}{(u)} ~ \mathrm{d} u$$

We note the integrand is even, as both $\cos{(u)}$ and $\mathrm{sech}{(u)}$ are even. Thus we have:

$$\frac{1}{2} \int_{-\infty}^{+\infty} \cos{(u)} \mathrm{sech}{(u)} ~ \mathrm{d} u = \int_{0}^{+\infty} \cos{(u)} \mathrm{sech}{(u)} ~ \mathrm{d} u$$

Now substitute $z =e^u$, so $u = \ln{z}$, and hence $\mathrm{d} u = \frac{1}{z} ~ \mathrm{d} z$. The bounds now become $e^0 = 1$ and $e^{+\infty} = + \infty$. Substituting:

$$\int_{0}^{+\infty} \cos{(u)} \mathrm{sech}{(u)} ~ \mathrm{d} u = \int_{1}^{+\infty} \cos{(\ln{z})} \mathrm{sech}{(\ln{z})} \cdot \frac{1}{z} ~ \mathrm{d} z$$

Using the complex exponential form of the hyperbolic secant, we see that:

$$\mathrm{sech}{(\ln{z})} = \frac{2z}{z^2 + 1}$$

And thus, our integral becomes:

$$\int_{1}^{+\infty} \frac{2z \cos{\ln{z}} }{z^2 + 1} \cdot \frac{1}{z} ~ \mathrm{d} z = 2 \int_{1}^{+\infty} \frac{\cos{\ln{z}} }{z^2 + 1} ~ \mathrm{d} z$$

We know from the complex exponential form of $\cos$ that:

$$\cos{\ln{z}} = \frac{1}{2} \left ( z^{-i} + z^i \right )$$

Substituting this into the integrand, we get:

$$\int_{1}^{+\infty} \frac{z^{-i} + z^i}{z^2 + 1} ~ \mathrm{d} z$$

And I can't seem to get any farther. I'll try and keep going later, I need sleep. Am I even on the right track, anyway?

Bacterius, my approach was a little different from yours. After using the substitution \(u=\ln(\tan x)\), I did

\[\begin{align} \int_0^\infty \cos(x) \text{sech}(x)dx &= 2\int_0^\infty \frac{e^{-x}}{1+e^{-2x}}\cos(x) dx\end{align}\]

Now use \(\displaystyle \frac{1}{1+e^{-2x}}=\sum_{n=0}^\infty (-1)^n e^{-2nx}\)

\[\begin{align} \int_0^\infty \cos(x)e^{-x} \sum_{n=0}^\infty (-1)^n e^{-2nx} dx &= \sum_{n=0}^\infty (-1)^n \int_0^\infty \cos(x) e^{-x(2n+1)}dx \\ &= \sum_{n=0}^{\infty}\frac{(-1)^n (2n+1)}{(2n+1)^2+1}\end{align}\]

This sum can be evaluated my many methods. Residues will be the easiest.

 

[ZaidAlyafey]

[tex]\int_{0}^{\frac{\pi}{2}}\cos(\ln(\tan(x)))dx[/tex]

[tex]\int^{\frac{\pi}{2}}_{0}\, \cos (\ln(\sin x )\,-\,\ln(\cos x ))\, dx[/tex]

[tex]=\int^{\frac{\pi}{2}}_{0}\, \cos (\ln(\sin x ))\,\cos(\ln(\cos x ))+\,\sin(\ln(\cos x ))\,\sin(\ln(\sin x ))\, dx[/tex]

[tex]\text{Which is the real part of : }(\sin x)^i \cdot (\cos x)^{-i}[/tex]

[tex]\int^{\frac{\pi}{2}}_0 \,(\sin x)^i \cdot (\cos x)^{-i} \, dx[/tex]

[tex]\text{Now I would use the so called property of beta function }[/tex]

[tex]\beta (x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta,[/tex]

[tex]\int^{\frac{\pi}{2}}_{0}\, \cos (\ln(\sin x ))\,\cos(\ln(\cos x ))+\,\sin(\ln(\cos x ))\,\sin(\ln(\sin x ))\, dx\,= \, \mathcal{Re} ( \int^{\frac{\pi}{2}}_{0} (\sin x)^i \cdot (\cos x)^{-i} ) [/tex]

[tex]\, \mathcal{Re}( \int^{\frac{\pi}{2}}_{0} (\sin x)^i \cdot (\cos x)^{-i} ) =\frac{1}{2} \, \mathcal{Re} ( \int^{\frac{\pi}{2}}_{0} (\sin x)^{2(\frac{1+i}{2})-1} \cdot (\cos x)^{2(\frac{1-i}{2})-1} ) \,=\,\frac{1}{2} \mathcal{Re}( \Gamma{(\frac{1+i}{2})} \, \Gamma{(\frac{1-i}{2})} )[/tex]

[tex]\frac{1}{2}\mathcal{Re}( \Gamma{(\frac{1+i}{2})} \, \Gamma{(\frac{1-i}{2})})=\frac{1}{2}\mathcal{Re}\( \Gamma{(\frac{1+i}{2})} \, \Gamma{(1-(\frac{1+i}{2}))}\)=\, \frac{1}{2}\,\mathcal{Re} ( \frac{\frac{\pi}{2}}{\sin(\frac{\pi+i\pi}{2})}) [/tex]

[tex] \sin(\frac{\pi+i\pi}{2})=\,\frac{e^{\frac{-\pi+i\pi}{2}}-e^{\frac{\pi-i\pi}{2}}}{2i}=\frac{ie^{\frac{-\pi}{2}}+ie^{\frac{\pi}{2}}}{2i}= \frac{e^{\frac{-\pi}{2}}+e^{\frac{\pi}{2}}}{2}=\cosh(\frac{\pi}{2}) [/tex]

[tex]\, \frac{1}{2}\,\mathcal{Re} ( \frac{\pi}{\sin(\frac{\pi+i\pi}{2})} ) = \frac{1}{2} \mathcal{Re}( \frac{\pi}{\cosh(\frac{\pi}{2})} ) = \frac {\pi}{2} \sec h ( \frac {\pi}{2})[/tex]

 

[sbhatnagar]

Very Good ZaidAlyafey!