Some problems with evaluating $\int_0^{\pi}\ln(\sin x+\sqrt{1+\sin^2x})dx$

Question

I'm trying to evaluate $$\int_{0}^{\pi}\ln\left(\sin\left(x\right) + \sqrt{\,1 + \sin^{2}\left(x\right)\,}\,\right)\,{\rm d}x . $$

I tried to evaluate it by making the substitution $u =\sin x$, but I faced a problem.

Since $\sin\left(\pi\right) = \sin\left(0\right) = 0$, the integral is equal to $0$. But by plotting the integrand, the integral seems to have a real non-zero value.

To avoid this problem I expressed the integral as $ \displaystyle 2\int_{0}^{\pi/2} \ln\left(\sin\left(x\right) + \sqrt{\,1 + \sin^{2}\left(x\right)\,}\,\right)\,{\rm d}x $. Then I tried letting $u = \sin x$.

How do you explain that the integral has a real non-zero value but by some substitution it's equal to zero?

And how does one evaluate the integral?

@MhenniBenghorbal I just try to find the area with arcsinh(sin x) — mnsh, Apr 11 '14 at 20:18
Your mistake is that $\sin(x)$ is not injective on $(0,\pi)$. — J.R., Apr 11 '14 at 20:22
@hammood arcsinh(sinx) is not the indefinite integral of your integrand. Remember that if $u=\sin x$ then $du=\cos x\ dx$; you're missing that $\cos x$ term in your integrand. — Steven Stadnicki, Apr 11 '14 at 20:27
oh yes I got it ,,,thanks .. now, what is the way to solve the integral ?????? — mnsh, Apr 11 '14 at 20:29
According to wolfram alpha the indefinite integrals closed form expression link — Alice Ryhl, Apr 11 '14 at 20:33
@Darksonn May I ask, how did you made wolfram alpha render such a big image? — chubakueno, Apr 11 '14 at 20:37
@chubakueno I just entered it into the box and it gave me that. I did it on the iPad app, that might make a difference — Alice Ryhl, Apr 11 '14 at 21:03

Random Variable · Accepted Answer · 2014-08-09T19:43:04.700

Using the Taylor expansion of $\text{arcsinh} (x)$ at $x=0$,

$$ \begin{align} \int_{0}^{\pi} \ln (\sin x + \sqrt{1+ \sin^{2} x}) \ dx &= \int_{0}^{\pi} \text{arcsinh}(\sin x) \ dx \\ &= 2 \int_{0}^{\pi/2} \text{arcsinh}(\sin x) \ dx \\ &= 2 \int_{0}^{\pi /2} \sum_{n=0}^{\infty} (-1)^{n} \binom{2n}{n} \frac{\sin^{2n+1}x}{2^{2n}(2n+1)} \ dx \\ &= 2 \sum_{n=0}^{\infty} (-1)^{n} \binom{2n}{n} \frac{1}{2^{2n}(2n+1)} \int_{0}^{\pi/2} \sin^{2n+1}(x) \ dx \\ &= 2 \sum_{n=0}^{\infty} (-1)^{n} \binom{2n}{n} \frac{1}{2^{2n}(2n+1)} \frac{1}{2n+1}\frac{2^{2n}}{\binom{2n}{n}} \tag{1} \\ &=2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = 2 G \end{align}$$

where $G$ is Catalan's constant.

$(1)$ The integral $ \displaystyle \int_{0}^{\pi /2} \sin^{2n+1} x \ dx $ can be evaluated by relating it to the beta function and then using the gamma duplication formula.

Specifically,

$$ \begin{align} \int_{0}^{\pi /2} \sin^{2n+1} x \ dx &= \int_{0}^{\pi/2} \sin^{2(n+1)-1} (x) \cos^{2(1/2)-1} (x) \ dx \\ &= \frac{1}{2} B \left(n+1,\frac{1}{2} \right) \\ &= \frac{1}{2} \frac{\Gamma(n+1) \Gamma(\frac{1}{2})}{\Gamma(n+\frac{3}{2})} = \frac{1}{2} \frac{\Gamma(n+1) \Gamma(\frac{1}{2})}{(n+\frac{1}{2})\Gamma(n+\frac{1}{2})} \\ &= \frac{\Gamma(n+1) \Gamma(\frac{1}{2})}{2n+1} \frac{2^{2n-1} \Gamma(n)}{\Gamma(2n) \Gamma(1/2)} \frac{2n}{2n} \\ &= \frac{1}{2n+1} 2^{2n} \frac{\Gamma(n+1) \Gamma(n+1)}{\Gamma(2n+1)} \\ &= \frac{1}{2n+1} \frac{2^{2n}}{\binom{2n}{n}} . \end{align} $$

score 0 · Answer 2 · answered Apr 11 '14 at 20:22

0

Looking the numerical result up in the Inverse Symbolic Calculator, it seems to be twice Catalan's constant

answered Apr 11 '14 at 20:22

Robert Israel

448,999

9

I think this would really fit as comment rather than as an answer.(the downvote isn't mine) – chubakueno Apr 11 '14 at 20:24

score 0 · Answer 3 · answered Jun 17 '14 at 22:22

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi}\ln\pars{\sin\pars{x} + \root{1 + \sin^{2}\pars{x}}}\,{\rm d}x :\ {\large ?}}$

\begin{align}&\color{#c00000}{% \int_{0}^{\pi}\ln\pars{\sin\pars{x} + \root{1 + \sin^{2}\pars{x}}}\,{\rm d}x} =2\int_{0}^{\pi/2}\ln\pars{\cos\pars{x} + \root{1 + \cos^{2}\pars{x}}}\,{\rm d}x \\[3mm]&=2\int_{0}^{\pi/2}{\rm arcsinh}\pars{\cos\pars{x}}\,{\rm d}x =2\int_{0}^{\pi/2}{\rm arcsinh}\pars{\sin\pars{x}}\,{\rm d}x \end{align}

However, $\ds{G = \int_{0}^{\pi/2}{\rm arcsinh}\pars{\sin\pars{x}}\,{\rm d}x}$ is a well known Catalan Constant $\ds{G}$ Integral Representation. See ${\bf\mbox{Entry}\ 17}$ in Victor Adamchik Page.

$$ \color{#66f}{\large% \int_{0}^{\pi}\ln\pars{\sin\pars{x} + \root{1 + \sin^{2}\pars{x}}}\,{\rm d}x = 2G} \approx 1.8319 $$

Some problems with evaluating $\int_0^{\pi}\ln(\sin x+\sqrt{1+\sin^2x})dx$

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