$\int_0^{\pi/2} \log^3 \cos x \, dx$
- Tolaso J Kos
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$\int_0^{\pi/2} \log^3 \cos x \, dx$
In continuity of this post by Grigorios let us see the same integral with a higher power. Prove that:
$$\int_0^{\pi/2} \log^3 \cos x\, {\rm d}x = - \frac{3\pi \zeta(3)}{4}- \frac{\pi^3 \log 2}{8} - \frac{\pi \log^3 2}{2}$$
$$\int_0^{\pi/2} \log^3 \cos x\, {\rm d}x = - \frac{3\pi \zeta(3)}{4}- \frac{\pi^3 \log 2}{8} - \frac{\pi \log^3 2}{2}$$
Imagination is much more important than knowledge.
Re: $\int_0^{\pi/2} \log^3 \cos x \, dx$
deleted.
Last edited by galactus on Sat Jan 23, 2016 1:41 pm, edited 1 time in total.
- Tolaso J Kos
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Re: $\int_0^{\pi/2} \log^3 \cos x \, dx$
Hey C,
well that's the main approach but let us see something different. I am starting from a formula Seraphim has presented us with, namely:
\begin{equation} \int_0^{\pi/2} (\log \cos x+ \log 2 + ix)^n \, {\rm d}x + \int_0^{\pi/2} (\log \cos x + \log 2 - ix)^n \, {\rm d}x=0 \end{equation}
Proof:
The function $f(z)=\log^n z $ is analytic in the open disk $\mathbb{D}=\{z \in \mathbb{C} \mid |z-1|<1\}$ since ($\mathfrak{Re}(1+z)>0$).
Using Gauss Mean Value Theorem we have that:
$$f(1)= \frac{1}{2\pi}\int_{0}^{2\pi} f\left ( 1+e^{ix} \right )\, {\rm d}x$$
However since $f(1)=0$ then we have successively:
\begin{align*}
0 &=\int_{0}^{2\pi} \log^n \left ( 1+\cos x +i \sin x \right )\, {\rm d}x \\
&= \int_{0}^{2\pi} \log^n \left ( 2\cos \frac{x}{2} e^{ix/2} \right )\, {\rm d}x\\
&\overset{u=x/2}{=\! =\! =\! =\!}\int_{0}^{\pi} \log^n \left ( 2\cos y \cdot e^{iy} \right )\, {\rm d}y \\
&= \int_{0}^{\pi/2} \log^n \left ( 2\cos x \cdot e^{iy} \right ) \, {\rm d}y + \int_{\pi/2}^{\pi} \log^n \left ( 2\cos y \cdot e^{iy} \right )\, {\rm d}y \\
&\overset{y =\pi-x }{=\! =\! =\! =\!} \int_{0}^{\pi/2}\log^n \left ( 2\cos x \cdot e^{ix} \right )\, {\rm d}y + \int_{0}^{\pi/2}\log^n \left ( -2 \cos x \cdot e^{i(\pi-x)} \right ) \, {\rm d}x \\
&= \int_{0}^{\pi/2}\log^n \left ( 2\cos x \cdot e^{ix} \right )\, {\rm d}x + \int_0^{\pi/2}\log^n \left ( 2\cos x \cdot e^{-ix} \right )\, {\rm d}x
\end{align*}
leading to the derivation of $(1)$. Note plug in $(1)$ $n=1, 2, 3$ to get the result I posted.
well that's the main approach but let us see something different. I am starting from a formula Seraphim has presented us with, namely:
\begin{equation} \int_0^{\pi/2} (\log \cos x+ \log 2 + ix)^n \, {\rm d}x + \int_0^{\pi/2} (\log \cos x + \log 2 - ix)^n \, {\rm d}x=0 \end{equation}
Proof:
The function $f(z)=\log^n z $ is analytic in the open disk $\mathbb{D}=\{z \in \mathbb{C} \mid |z-1|<1\}$ since ($\mathfrak{Re}(1+z)>0$).
Using Gauss Mean Value Theorem we have that:
$$f(1)= \frac{1}{2\pi}\int_{0}^{2\pi} f\left ( 1+e^{ix} \right )\, {\rm d}x$$
However since $f(1)=0$ then we have successively:
\begin{align*}
0 &=\int_{0}^{2\pi} \log^n \left ( 1+\cos x +i \sin x \right )\, {\rm d}x \\
&= \int_{0}^{2\pi} \log^n \left ( 2\cos \frac{x}{2} e^{ix/2} \right )\, {\rm d}x\\
&\overset{u=x/2}{=\! =\! =\! =\!}\int_{0}^{\pi} \log^n \left ( 2\cos y \cdot e^{iy} \right )\, {\rm d}y \\
&= \int_{0}^{\pi/2} \log^n \left ( 2\cos x \cdot e^{iy} \right ) \, {\rm d}y + \int_{\pi/2}^{\pi} \log^n \left ( 2\cos y \cdot e^{iy} \right )\, {\rm d}y \\
&\overset{y =\pi-x }{=\! =\! =\! =\!} \int_{0}^{\pi/2}\log^n \left ( 2\cos x \cdot e^{ix} \right )\, {\rm d}y + \int_{0}^{\pi/2}\log^n \left ( -2 \cos x \cdot e^{i(\pi-x)} \right ) \, {\rm d}x \\
&= \int_{0}^{\pi/2}\log^n \left ( 2\cos x \cdot e^{ix} \right )\, {\rm d}x + \int_0^{\pi/2}\log^n \left ( 2\cos x \cdot e^{-ix} \right )\, {\rm d}x
\end{align*}
leading to the derivation of $(1)$. Note plug in $(1)$ $n=1, 2, 3$ to get the result I posted.
Imagination is much more important than knowledge.
Re: $\int_0^{\pi/2} \log^3 \cos x \, dx$
I am gonna delete my post. It's too cliche'. I want to do it differently.
That is cool. Leave to ol' Seraphim:)
When n=3, doesn't it lead to something maybe easier to deal with:
$$\int_{0}^{\frac{\pi}{2}}\ln^{3}(2\cos(x))=\int_{0}^{\frac{\pi}{2}}3x^{2}\ln(2\cos(x))$$
That's a clever approach, T.
EDIT:
You know, considering the idea above, it is easier to consider $$3\int_{0}^{\frac{\pi}{2}}x^{2}\ln(2\cos(x))dx$$, then the original cubic.
I thought I may try contours. There is a contour approach for these I learned a good while ago from someone on SE.
Use a rectangular contour with quarter-circle indents at $\pm \frac{\pi}{2}$ and with vertices:
$$\pm \frac{\pi}{2}, \;\ \pm \frac{\pi}{2}+Ri$$
and consider $$f(z)=z^{2}\log(1+e^{2iz})$$
This is related to what you have above, T.
$$\frac{1}{2}\cdot 3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x^{2}(\log(2\cos(x))+ix)dx=3i/2\int_{0}^{\infty}[(iy+\frac{\pi}{2})^{2}-(iy-\frac{\pi}{2})^{2}]\log(1-e^{-2y})dy$$
$$i[f(\frac{\pi}{2}+iy)-f(\frac{\pi}{2}-iy)]=2\pi y$$
Thus, we have $$-3\pi \int_{0}^{\infty}y\log(1-e^{-2y})dy$$
$$-3\pi \int_{0}^{\infty}y\sum_{n=1}^{\infty}\frac{e^{-2ny}}{n}dy$$
$$=-3\pi \sum_{n=1}^{\infty}\int_{0}^{\infty}ye^{-2ny}dy=\frac{-3\pi}{4}\zeta(3)$$
That is cool. Leave to ol' Seraphim:)
When n=3, doesn't it lead to something maybe easier to deal with:
$$\int_{0}^{\frac{\pi}{2}}\ln^{3}(2\cos(x))=\int_{0}^{\frac{\pi}{2}}3x^{2}\ln(2\cos(x))$$
That's a clever approach, T.
EDIT:
You know, considering the idea above, it is easier to consider $$3\int_{0}^{\frac{\pi}{2}}x^{2}\ln(2\cos(x))dx$$, then the original cubic.
I thought I may try contours. There is a contour approach for these I learned a good while ago from someone on SE.
Use a rectangular contour with quarter-circle indents at $\pm \frac{\pi}{2}$ and with vertices:
$$\pm \frac{\pi}{2}, \;\ \pm \frac{\pi}{2}+Ri$$
and consider $$f(z)=z^{2}\log(1+e^{2iz})$$
This is related to what you have above, T.
$$\frac{1}{2}\cdot 3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x^{2}(\log(2\cos(x))+ix)dx=3i/2\int_{0}^{\infty}[(iy+\frac{\pi}{2})^{2}-(iy-\frac{\pi}{2})^{2}]\log(1-e^{-2y})dy$$
$$i[f(\frac{\pi}{2}+iy)-f(\frac{\pi}{2}-iy)]=2\pi y$$
Thus, we have $$-3\pi \int_{0}^{\infty}y\log(1-e^{-2y})dy$$
$$-3\pi \int_{0}^{\infty}y\sum_{n=1}^{\infty}\frac{e^{-2ny}}{n}dy$$
$$=-3\pi \sum_{n=1}^{\infty}\int_{0}^{\infty}ye^{-2ny}dy=\frac{-3\pi}{4}\zeta(3)$$
- Tolaso J Kos
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Re: $\int_0^{\pi/2} \log^3 \cos x \, dx$
Besides, the above approach with contours here is another with real analysis, since it always transmits something magical.galactus wrote:
When $n=3$, doesn't it lead to something maybe easier to deal with:
$$\int_{0}^{\frac{\pi}{2}}\ln^{3}(2\cos(x))=\int_{0}^{\frac{\pi}{2}}3x^{2}\ln(2\cos(x))$$
That's a clever approach, T.
Recalling the Fourier $\cos $ series of $\log \cos x$ we have that:
\begin{equation} \ln \cos x = -\ln 2 +\sum_{n=1}^{\infty}\frac{(-1)^{n-1} \cos 2n x}{n} \end{equation}
we have that:
\begin{align*}
\int_{0}^{\pi/2} \ln^3 2\cos x \, {\rm d}x &=3\int_{0}^{\pi/2}x^2 \ln 2 \cos x \, {\rm d}x \\
&\overset{(2)}{=} 3\int_{0}^{\pi/2}x^2 \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cos 2nx}{n} \, {\rm d}x\\
&=3\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_{0}^{\pi/2}x^2 \cos 2n x \, {\rm d}x \\
&=- \frac{3\pi}{4}\sum_{n=1}^{\infty}\frac{1}{n^3} \\
&=-\frac{3\pi \zeta(3)}{4}
\end{align*}
Using the above formula and gathering terms we get the result.
P.S: C, your answer might have been cliché, but there was no need to delete it.
Imagination is much more important than knowledge.
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Re: $\int_0^{\pi/2} \log^3 \cos x \, dx$
$$I=\int\limits_{0}^{\frac{\pi }{2}}{\log ^{3}\cos xdx}=\int\limits_{0}^{\frac{\pi }{2}}{\sin ^{2\cdot \frac{1}{2}-1}x\cos ^{2\cdot \frac{1}{2}-1}x\log ^{3}\cos xdx}$$Tolaso J Kos wrote:In continuity of this post by Grigorios let us see the same integral with a higher power. Prove that:
$$\int_0^{\pi/2} \log^3 \cos x\, {\rm d}x = - \frac{3\pi \zeta(3)}{4}- \frac{\pi^3 \log 2}{8} - \frac{\pi \log^3 2}{2}$$
$$=\frac{\partial ^{3}}{\partial v^{3}}\left( \int\limits_{0}^{\frac{\pi }{2}}{\sin ^{2u-1}x\cos ^{2v-1}xdx} \right)\left| _{\left( x,y \right)=\left( \frac{1}{2},\frac{1}{2} \right)} \right.=\frac{1}{16}\cdot \frac{\partial ^{3}\beta \left( u,v \right)}{\partial v^{3}}\left| _{\left( u,v \right)=\left( \frac{1}{2},\frac{1}{2} \right)} \right.$$
$$=\frac{1}{16}\left( \left( \psi ^{\left( 0 \right)}\left( v \right)-\psi ^{\left( 0 \right)}\left( u+v \right) \right)^{3}+3\left( \psi ^{\left( 1 \right)}\left( v \right)-\psi ^{\left( 1 \right)}\left( u+v \right) \right)\left( \psi ^{\left( 0 \right)}\left( v \right)-\psi ^{\left( 0 \right)}\left( u+v \right) \right)+\psi ^{\left( 2 \right)}\left( v \right)-\psi ^{\left( 2 \right)}\left( u+v \right) \right)\beta \left( u,v \right)\left| _{\left( x,y \right)=\left( \frac{1}{2},\frac{1}{2} \right)} \right.$$
$$=\frac{1}{16}\left( \left( \psi ^{\left( 0 \right)}\left( \frac{1}{2} \right)-\psi ^{\left( 0 \right)}\left( 1 \right) \right)^{3}+3\left( \psi ^{\left( 1 \right)}\left( \frac{1}{2} \right)-\psi ^{\left( 1 \right)}\left( 1 \right) \right)\left( \psi ^{\left( 0 \right)}\left( \frac{1}{2} \right)-\psi ^{\left( 0 \right)}\left( 1 \right) \right)+\psi ^{\left( 2 \right)}\left( \frac{1}{2} \right)-\psi ^{\left( 2 \right)}\left( 1 \right) \right)\beta \left( \frac{1}{2},\frac{1}{2} \right)$$
$$=\frac{1}{16}\left( \left( -\gamma -2\log 2+\gamma \right)^{3}+3\left( \frac{\pi ^{2}}{2}-\frac{\pi ^{2}}{6} \right)\left( -\gamma -2\log 2+\gamma \right)-14\zeta \left( 3 \right)-\left( -2\zeta \left( 3 \right) \right) \right)\pi $$
$$=-\left( \frac{\pi }{2}\log ^{3}2+\frac{\pi ^{3}}{8}\log 2+\frac{3\pi }{4}\zeta \left( 3 \right) \right)$$
Regards
Civil Engineer
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