Show that $\int_0^{\frac\pi 2}\frac{\theta-\cos\theta\sin\theta}{2\sin^3\theta}d\theta=\frac{2C+1}4$

Question

While trying to evaluate $\int_0^1 k^2K(k)dk$ related to elliptic integral of the first kind, by integral switching method, I reached the trigonometric integral $$\int_0^{\frac\pi 2}\frac{\theta-\cos\theta\sin\theta}{2\sin^3\theta}d\theta$$ which is evaluated to $\frac{2C+1}4$ by Wolfram Alpha. Here, $C$ is the Catalan constant, sometimes denoted by $G$. This integral is complicated for me.

Are there another methods to evaluate the integrals $\int_0^1k^nK(k)dk$, $n\geq2$? Or can someone please evaluate the trigonometric integral above? Thank you.

Answer 1

With $K$ and $E$ as the Complete Elliptic Integral of the First and Second Kind respectively with parameter $k$.

Denote

$$K_n=\int_0^1k^nK\ dk$$

$$E_n=\int_0^1k^nE\ dk$$

then one can prove that,

$$n^2K_n-(n-1)^2K_{n-2}=1\tag{1}$$

with the initial values of

$$K_0=2G,\quad K_1=1$$

One may have the following closed forms by building upon the recurrence,

$$\int_{0}^{1}k^{2n+1}K\ dk=\frac{2^{4n}}{\left(2n+1\right)^{2}}\binom{2n}{n}^{-2}\sum_{k=0}^{n}\binom{2k}{k}^2\frac{1}{2^{4k}}$$

$$\int_{0}^{1}k^{2n}K\ dk=\frac{1}{2^{4n}}\binom{2n}{n}^2\left[2G+\frac{1}{4}\sum_{k=1}^n\frac{2^{4k}}{k^2\displaystyle\binom{2k}{k}^2}\right]$$

The proof of $(1)$ is as follows

Use IBP, Integrate $kK$ and differentiate $k^{n-1}$

Using the following result $$\int kK\ dk=E-(1-k^2)K$$

$$K_n=1-(n-1)\int_0^1k^{n-2}[E-(1-k^2)K]\ dk\tag{2}$$

Now we need relation between $E_n$ and $K_n$

One can use the Derivative of $E$ and obtain

$$[k^nE]'=k^{n-1}[E-K]+nk^{n-1}E$$

Then integrate to obtain $$1+K_n=(n+2)E_n\tag{3}$$

Using $(2)$ and $(3)$ the result in $(1)$ follows.

Answer 2

Hint

If you let $t=2 \tan ^{-1}(x)$, the antiderivative becomes $$4I=\int\frac{x^2-1}{x^2}\,dx+\int \frac{\left(x^2+1\right)^2 \tan ^{-1}(x)}{x^3}\,dx$$ Expanding $$4I=\int\Bigg(1-\frac{1}{x^2} +x \tan ^{-1}(x)+2\frac{ \tan ^{-1}(x)}{x}+\frac{ \tan ^{-1}(x)}{x^3}\Bigg)\,dx$$

Use the logarithmic representation of the arctangent function and a couple of integration by parts to get the result.

If you prefer to use series $$I=-2 \sum_{n=0}^\infty \int \frac {(-1)^n}{(2 n-1) (2 n+1) (2 n+3) }\,x^{2n}\,dx$$ $$\int_0^1\frac{x^2-1}{x^2}\,dx+\int \frac{\left(x^2+1\right)^2 \tan ^{-1}(x)}{x^3}\,dx=-2 \sum_{n=0}^\infty \frac {(-1)^n}{(2 n-1) (2 n+1)^2 (2 n+3) }$$ and the result.

Answer 3

Integrate the function $$f(z) = \frac{z-\cos(z)\sin(z) }{2\sin^{3}(z)} $$ around a tall rectangular contour with vertices at $0$, $\pi/2$, $\pi/2+iR$, and $iR$.

As $R \to \infty$, the integral vanishes on the top of the rectangle because the magnitude of $f(z)$ decays exponentially to zero as $\Im(z) \to \infty$.

And the integral on the left side of the rectangle is purely imaginary.

Since there are no poles inside the contour, we have $$ \begin{align} \int_{0}^{\pi /2} \frac{\theta-\cos(\theta) \sin(\theta)}{2 \sin^{3}(\theta)} \, \mathrm d \theta &= - \Re \int_{0}^{\infty} \frac{\tfrac{\pi}{2}+it - \cos (\tfrac{\pi}{2}+it) \sin(\tfrac{\pi}{2}+it)}{2 \sin^{3} (\frac{\pi}{2}+it)} \, i \, \mathrm dt \\ &= - \Re \int_{0}^{\infty} \frac{\tfrac{\pi}{2} + it+ i \sinh(t) \cosh(t)}{2 \cosh^{3}(t)} \, i \, \mathrm dt \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{t}{\cosh^{3}(t)} \, \mathrm dt + \frac{1}{2} \int_{0}^{\infty} \frac{\sinh(t)}{\cosh^{2}(t)} \, \mathrm dt \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{t}{\cosh^{3}(t)} \, \mathrm dt + \frac{1}{2}. \end{align}$$

An integral representation for the Dirichlet beta function is $$\beta(s) = \frac{1}{2 \Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{\cosh t} \, \mathrm dt, \quad \Re(s) >0. \tag{1}$$

Integrating by parts twice, we get the equation $$ \begin{align} \beta(s) &= \frac{1}{2 \Gamma(s+2)} \left(\int_{0}^{\infty} \frac{t^{s+1}}{\cosh(t)} \, \mathrm dt- 2\int_{0}^{\infty}\frac{t^{s+1}}{\cosh^{3}(t)} \, \mathrm dt \right) \\ &= \frac{1}{2 \Gamma(s+2)} \left(2 \Gamma(s+2) \beta(s+2)- 2\int_{0}^{\infty}\frac{t^{s+1}}{\cosh^{3}(t)} \, \mathrm dt\right) \\ &= \beta(s+2) - \frac{1}{\Gamma(s+2)}\int_{0}^{\infty}\frac{t^{s+1}}{\cosh^{3}(t)} \, \mathrm dt, \end{align}$$ which holds for $\Re(s) >-2$.

Therefore, $$\int_{0}^{\infty} \frac{t}{\cosh^{3}(t)} \, \mathrm dt = \beta(2) - \beta(0) \overset{\spadesuit}{=} G- \frac{1}{2}, $$ and $$ \int_{0}^{\pi /2} \frac{\theta-\cos(\theta) \sin(\theta)}{2 \sin^{3}(\theta)} \, \mathrm d \theta = \frac{1}{2} \left(G- \frac{1}{2} \right) + \frac{1}{2}= \frac{G}{2}+ \frac{1}{4}.$$

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$\spadesuit$ The value of the Dirichlet beta function at zero can be obtained from the functional equation, or it can be obtained from the integral representation $$\beta(s) = \frac{1}{2 \Gamma(s+1)} \int_{0}^{\infty} \frac{t^{s}\sinh(t)}{\cosh^{2}(t)} \, \mathrm dt , \quad \Re(s) >-1, $$ which itself can be obtained by integrating $(1)$ by parts.

Answer 4

With $\int_0^{\frac\pi 2}\frac{t}{\sin t}dt= 2G $ \begin{align} &\int_0^{\frac\pi 2}\frac{t-\cos t\sin t}{\sin^3t}dt\\ = &\int_0^{\frac\pi 2}\left(\frac{t}{2\sin t} +(t-\cos t\sin t)\frac{2-\sin^2t}{2\sin^3t}-\frac12\cos t\right) dt\\ = &\ G +\int_0^{\frac\pi 2} \frac{t-\cos t\sin t}2 \overset{ibp} d\left(-\frac{\cos t}{\sin^2t}\right)-\frac12\cos t dt\\ =& \ G +\frac12 \int_0^{\frac\pi 2}\cos t dt= G+\frac12 \end{align}