An extension of "turn" Euler's integral

Solve $$\int_0^{\frac{\pi}{2}} {{{\ln}^2}\sin XDX} = \int_0^{\frac{\pi}{2}} {{{\ln}^2}\cos XDX} $$ value.

Apparently \[\int_0^{\frac{\pi}{2}} {{{\ln}^2}\sin XDX} \underline{\underline {{\text{order}x = \frac{\pi}{2}-t}}} \int_0^{\frac{\ Pi}{2}} {{{\ln}^2}\cos xdx}.\]

Remember to ask for integral for $i$, notice

(1) $\int_0^\pi {{\ln}^2}\sin \left (x \right) dx} = 2\int_0^{\frac{\pi}{2}} {{{\ln}^2}\sin XDX} = 2i$;

(2) $\int_0^\pi {\ln \sin XDX} = 2\int_0^{\frac{\pi}{2}} {\ln \sin XDX} \text{(Euler integral)} =-\pi \ln 2$;

(3) $\sum\limits_{n = 1}^\infty {\frac{{{{\left ({-1} \right)}^{n-1}}}}{{{{\left ({2n-1} \right)}^3}}}}=\beta (3) = \frac{{{\pi ^3}}}{32}$, where $\beta (s) $ is Dirichlet beta function;

(4) $\int_0^{\frac{\pi}{2}} {{{\ln}^2}\tan XDX} = \int_0^{\frac{\pi}{2}} {{{\ln}^2}\cot XDX} = \frac{{{\pi ^3}}}{8}.$

In fact, we have

\begin{align*}\int_0^{\frac{\pi}{2}} {{{\ln}^2}\tan XDX} &= \int_0^\infty {\frac{{{{\ln}^2}x}}{{1 + {X^2}}}DX} = 2\int_0^1 {\frac{{{{\ln}^2}x}}{{1 + {X^2}}}DX} = 2\int_0^1 {{{\ln}^2}x\sum\limits_{n = 1}^\infty {{{\left}} \right)}^{n-1}}} dx} \\&= 2\sum\limits_{n = 1}^\infty {{{\left ({-1} \right)}^{n-1}}\int_0^1 {{x^{2n-2}}{{\l   N}^2}XDX}} = 4\sum\limits_{n = 1}^\infty {\frac{{{{\left ({-1} \right)}^{n-1}}}}{{{{\left ({2n-1} \right)}^3}}} = \frac{{{\pi ^3}}}{8}.\end{align*}

and

\begin{align*}2i &= \int_0^{\frac{\pi}{2}} {{\left[\left {\ln ({2x} \sin)-\left 2} \right)} \ln]} ^2}DX}-\int_0^{\frac{\pi}{2}} {2\ln \sin x\ln \cos XDX} \\&{= \int_0^{\frac{\pi}{2}} {{{\ln}^2}\sin \left ({2x} \right) dx}-2\ln 2\int_0^{\frac{\pi}{2}} {{\rm{lnsin}}\left ({2x} \right) dx} + \frac{{{{\ln}^2}2}}{2}\pi-\int_0^{ \frac{\pi}{2}} {2\ln \sin x\ln \cos XDX}}\\&{= \frac{1}{2}\int_0^\pi {{{\ln}^2}\sin \left (x \right) dx}-\ln 2\ int_0^\pi {\ln \sin XDX} + \frac{{{{\ln}^2}2}}{2}\pi-\int_0^{\frac{\pi}{2}} {2\ln \sin x\ln \cos XDX}}\\&{= I + \frac{{3{{\ln}^2}2}}{2}\pi-2\int_0^{\frac{\pi}{2}} {\ln \sin x\ln \cos XDX}}.\end{align*}

\[2i = \int_0^{\frac{\pi}{2}} {{{\ln}^2}\tan XDX} + 2\int_0^{\frac{\pi}{2}} {\ln \sin x\ln \cos XDX}. \]

So

\begin{align*}3i &= \int_0^{\frac{\pi}{2}} {{{\ln}^2}\tan XDX} + \frac{{3{{\ln}^2}2}}{2}\pi = \frac{{{\pi ^3}}}{8 } + \frac{{3{{\ln}^2}2}}{2}\pi \\i &= \frac{{{\pi ^3}}}{{24} + \frac{{{{\ln}^2}2}}{2}\pi.\end{align*}

At the same time we have

\[\int_0^{\frac{\pi}{2}} {\ln \sin x\ln \cos XDX} = \frac{{{{\ln}^2}2}}{2}\pi-\frac{{{\pi ^3}}}{{48}}.\]

This question is very good, haha, is an extension of Euler integral.

For Dirichlet beta functions, refer to:

Http://mathworld.wolfram.com/DirichletBetaFunction.html or Http://en.wikipedia.org/wiki/Dirichlet_beta_function.

Solution Two: $$ \beta (x, y) = 2\int_0^{\frac{\pi}{2}} \sin^{2x-1} (t) \cos^{2y-1} (t) \ dt $$

$$ \frac{\partial}{\partial x} \beta (x, y) = 4\int_0^{\frac{\pi}{2}} \sin^{2x-1} (t) \ln (\sin t) \cos^{2y-1} (t) \ dt$$

$$ \frac{\partial}{\partial y} \left (\frac{\partial}{\partial x} \beta (x, y) \right) = 8\int_0^{\frac{\pi}{2}} \sin^{ 2x-1} (t) \ln (\sin t) \ln (\cos t) \cos^{2y-1} (t) \ dt $$

And we have $$ \beta (x, y) = \frac{\gamma (×) \gamma (y)}{\gamma (x+y)} $$

So differentiate and put $ x =\frac{1}{2}, y = \frac{1}{2} $

$$ \psi \left (\frac{1}{2} \right) = -2\ln 2-\gamma $$

$$ \psi (1) =-\gamma $$

$$ \psi^{(1)} (1) = \frac{\pi^2}{6} $$

$$ \beta \left (\frac{1}{2}, \frac{1}{2} \right) = \frac{\gamma \left (\frac{1}{2} \right) ^2}{\gamma (1)} = \pi $$

Thus you'll have the integral $ = \frac{\pi}{8} \left (4\ln (2) ^2-\frac{\pi^2}{6} \right) $

See: Http://math.stackexchange.com/questions/492878/find-int-0-frac-pi2-ln-sinx-ln-cosx-mathrm-dx?rq=1

Solution Three: A third approach would be the the Fourier series:

Namely, consider

$$\ln \left (2\sin \frac{x}{2}\right) =-\sum_{n=1}^{\infty}\frac{\cos nx}{n};(0<x<2\pi) $$

After squared:

$$\ln^2\left (2\sin \frac{x}{2}\right) =\sum_{n=1}^{\infty} \sum_{k=1}^{\infty}\frac{\cos Kx\cos nx}{kn}$$

Now, integrate the last equation from $x =0$ to $x =\pi$

On the right side, we get:

$$\frac{\pi}{2}\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi}{2}\frac{\pi^2}{6}=\frac{\pi^3}{12}$$

Because $ $I =\int_{0}^{\pi}\cos kx\cos nx\,dx=0;k\neq n$$ $ $I =\frac{\pi}{2};k=n$$

On the left side:

$$\int_{0}^{\pi}\ln^2\left (2\sin \frac{x}{2}\right) \,dx= \ln^22 \int_{0}^{\pi}dx + 4\ln 2 \int_{0}^{\frac{\pi}{2}} \LN \left (\sin x\right) dx+2 \int_{0}^{\frac{\pi}{2}} \ln^2 \left (\sin x\right) dx $$

Since we know that $ \int_{0}^{\frac{\pi}{2}} \ln (\sin x) dx=-\frac{\pi}{2}\ln (2) $ then we get $ \int_{0}^{\frac{\pi}{2}} \LN^{2} (\sin x) dx $ from the equation.

Solution Four:

Here is a completely different-approach this integral, which relies on some elementary complexanalysis ( cauchy ' s theorem). It is based in an approach which I had seen several times employed to compute $\int_0^{\pi/2}\log{(\sin{x})}\,dx$ (partic Ularly, in Ahlfors's book).

The idea was to integrate the principal branch of $f (z): = \log^2{(1-e^{2iz})} = \log^2{( -2ie^{iz}\sin{z})}$ over the con Tour below, and then let $R \to \infty$ and $\epsilon \to 0$.

First of all, $1-e^{2iz} \leq 0$ If $z = K\pi + iy$, where $k $ is integral and $y \leq 0$. Thus, in the region of the plane which are obtained by omitting the lines $\{k\pi + iy:y\leq 0\}$ for $k \in \mathbb z$, t He principal branch of $\log{(1-e^{2iz})}$ is defined and analytic. Note that, for each fixed $R $ and $\epsilon$, the contour we wish to integrate over are contained entirely within this Regi On.

By Cauchy ' s theorem, the integral over the contour vanishes for each fixed $R $. Since $f (x + IR) = \log^2{(1-e^{2ix}e^{-2r})} \to 0$ uniformly as $R \to \infty$, the integral over the segment $[IR,\PI /2 + ir]$ vanishes in the limit. Similarly, since $1-e^{2iz} = O (z) $ as $z \to 0$, we have $f (z) = O (\log^2{|z|}) $ for small enough $z $, which, since $\epsilon \log^2{\epsilon} \to 0$ with $\epsilon$, means that the the integral over The circular arc from $i \epsilon$ to $\epsilon$ vanishes as $\epsilon \to 0$.

From the vertical sides of the contour, we get the the contribution

\BEGIN{ALIGN*}\INT_{[\PI/2,\PI/2 + IR]} + \int_{[ir,i\epsilon]} f (z) \,dz & = I\int_0^r F (\pi/2 + iy) \,dy-i\int_\epsil On^r f (iy) \,dy.\end{align*}

Since $f (iy) $ and $f (\PI/2 + iy) $ is real, this contribution is purely imaginary.

Finally, the contribution from the bottom side of the contour, after letting $\epsilon \to 0$, is

\BEGIN{ALIGN*}\INT_0^{\PI/2} f (x) \,dx = \INT_0^{\PI/2} \log^2{( -2ie^{ix}\sin{x})}\,dx,\end{align*}

And we know from the preceding remarks, the real part of this integral must vanish. For $x $ between $0$ and $\pi/2$, the quantity $2\sin{x}$ is positive. Writing $-ie^{ix} = e^{i (x-\PI/2)}$, we see this $x-\pi/2$ is the unique value of $\arg{( -2ie^{ix}\sin{x})}$ which lie S in $ (-\PI,\PI) $. Since we have chosen the principal branch of $\log{z}$, it follows from these considerations that $\log{( -2ie^{ix}\sin{x}) } = \log{(2\sin{x})} + I (X-\PI/2) $, and therefore that

\begin{align*}\text{re}{f (x)} &= \log^2{(2\sin{x})}-(X-\PI/2) ^2 \\&= \log^2{(\sin{x})} + 2\log{2}\log{(\sin{ X})} + \log^2{2}-(X-\PI/2) ^2.\end{align*}

By setting $\INT_0^{\PI/2} \text{re}f (x) \,dx = 0$ we get

\BEGIN{ALIGN*}\INT_0^{\PI/2} \log^2{(\sin{x})}\,dx &= \INT_0^{\PI/2} (X-\PI/2) ^2\,DX-2\LOG{2}\INT_0^{\PI/2} \ log{(\sin{x})}\,dx-\frac{\pi}{2}\log^2{2} \\& = \frac{1}{3}\left (\frac{\pi}{2}\right) ^3 + \frac{\pi}{2} \log^2{2 }\end{align*}

As expected

By similar methods, one can compute a variety of integrals of this form with little difficulty. Here is some examples I has computed for fun. All is proved by the same method, with the same contour, but different functions $f $.

1. Take $f (z) = \log{(1 + e^{2iz})} = \log{(2e^{iz}\cos{z})}$ and compare imaginary parts to get

$$\int_0^\infty \log{(\coth{y})}\,dy = \frac{1}{2}\left (\frac{\pi}{2}\right) ^2.$$

2. Related to the question of yours (which incidentally led me here), one can show by taking $f (z) = \log^4 (1 + e^{2iz}) $ And comparing real parts that

$$\INT_0^{\PI/2} x^2\log^2{(2\cos{x})}\,dx = \frac{1}{30}\left (\frac{\pi}{2}\right) ^5 + \FRAC{1}{6}\INT_0^{\PI/2} \ log^4{(2\cos{x})}\,dx.$ $Assuming The result of the other question, we then get

$$\INT_0^{\PI/2} \log^4{(2\cos{x})}\,dx = \frac{19}{15}\left (\frac{\pi}{2}\right) ^5.$$

3.Also related to the question cited in 2., taking $f (z) = z^2\log^2{(1 + e^{2iz})}$ and comparing real parts gives

$$\int_0^{\pi/2}x^2\log^2{(2\cos{x})}\,dx = \frac{1}{5}\left (\frac{\pi}{2}\right) ^5 + \pi \int_0^\infty y\log^2{(1-e ^{-2Y})}\,dy.$$

Once more, assuming the result of the other question, we get

$$\int_0^\infty y\log^2{(1-e^{-2y})}\,dy = \frac{1}{45}\left (\frac{\pi}{2}\right) ^4.$$

Actually, the integral in 3. Have several interesting series expansions, and I would be very interested if someone could compute it without using the RE Sult from the question I cited. For one thing, that would give us a different proof of this result (which is what I started investigating it in the first P Lace).

See: http://math.stackexchange.com/questions/58654/ integrate-square-of-the-log-sine-integral-int-0-frac-pi2-ln2-sinx/58672#58672

Then look at the promotion:

\[\int_0^{\frac{\pi}{2}} {{{\ln}^n}\sin XDX} = \frac{{\sqrt \pi}}{{{2^{n + 1}}}}\frac{{{d^k}}}{{d{\alpha ^k}}}{\left ( {\frac{{\gamma \left ({\frac{{2\alpha + 1}}{2}} \right)}}{{\gamma \left ({\frac{{2\alpha + 2}}{2}} \right)}}} \right) _{ \alpha = 0}}.\]

proof. Note \[\int_0^{\frac{\pi}{2}} {{{\ln}^n}\sin XDX} = \int_0^{\frac{\pi}{2}} {\frac{{{{\ln}^n}t}}{{\sqrt {1-{t^2}}}}DT }. \]

Make \[i\left (\alpha \right) = \int_0^1 {\frac{{{t^{2\alpha}}}}{{\sqrt {1-{t^2}}}}dt} = \frac{1}{2}b\left ({\frac{1}{2} , \frac{{2\alpha + 1}}{2}} \right) = \frac{{\sqrt \pi}}{2}\frac{{\gamma \left ({\frac{{2\alpha + 1}}{2}} \right)}}{{\Ga MMA \left ({\frac{{2\alpha + 2}}{2}} \right)}},\]

Then there is \[{i^{\left (n \right)}}\left (\alpha \right) = \int_0^1 {\frac{{{t^{2\alpha}}{2^n}{{\ln}^n}t}}{{\sqrt {1-{t^2}}} DT}. \]

Therefore there are

\[\int_0^{\frac{\pi}{2}} {{{{\ln}^n}\sin XDX} = \int_0^{\frac{\pi}{2}} {\frac{{{{\ln}^n}t}}{{\sqrt {1-{t^2}}}}dt} = \frac{{{i^{\left (n \right)}}\left (0 \right)}}{{{2^n}} = \frac{{\sqrt \pi}}{{{2^{n + 1}}}}\frac{{{d^k}}}{{d{\alpha ^ K}}}{\left ({\frac{{\gamma \left ({\frac{{2\alpha + 1}}{2}} \right)}}{{\gamma \left ({\frac{{2\alpha + 2}}{2}} \right)}} } \right) _{\alpha = 0}}.\]

An alternative solution

Use generating series. Consider the function $ $f (z) =\sum_{k=0}^{\infty}s_{k}\frac{z^{k}}{k!}. $$ then $ $f (z) =\int_0^1 \left (\sum_{k=0}^\infty ( -1) ^k \log^k (\sin (\pi x)) z^k \right) dx=\int_{0}^{1}\frac{1}{\sin\  Left (\pi x\right) ^{z}}dx=\frac{\gamma\left (\frac{1-z}{2}\right)}{\sqrt{\pi}\gamma\left (1-\frac{z}{2}\right)}.$$ The last equality follows from a identity of the Beta Function and then applying the duplication Formula. From here, differentiating and plugging in $z =0$ allows you to conclude.

See: Http://math.stackexchange.com/questions/121473/solve-the-integral-s-k-1k-int-01-log-sin-pi-xk-dx?rq=1

You can also refer to these questions:

Http://math.stackexchange.com/questions/307593/a-hard-log-definite-integral-int-0-pi-4-ln3-sin-x-mathrm-dx?lq=1

or http://math.stackexchange.com/questions/289587/log-sin-and-log-cos-integral-maybe-relate-to-fourier-series/309781#309781.

Http://math.stackexchange.com/questions/330057/how-to-evaluate-i-displaystyle-int-0-pi-2x2-ln-sin-x-ln-cos-xdx

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An extension of "turn" Euler's integral