$$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx \\|b|<1$$

I tried putting $-x$ using properties of definite integral, but that doesn't really help.

$$I=\int_{-\frac\pi2}^{\frac\pi2}\frac{-\ln(1-b\sin x)}{\sin x}dx$$

I don't think adding these 2 equations would yield anything. And I just cannot think of a good substitution.

Edit : Plugging b=1, it gives 4.9xxx . as $\pi ^2 \approx 10$, the value is $\pi \arcsin b$. Other values for b confirm this.

  • $\begingroup$ Maybe integration by parts? $\endgroup$
    – orion
    Mar 27, 2014 at 8:06
  • $\begingroup$ @orion I think that it will complicate it further. $\endgroup$
    – evil999man
    Mar 27, 2014 at 8:09
  • $\begingroup$ Even wolfram alpha isn't replying... $\endgroup$
    – evil999man
    Mar 27, 2014 at 8:30
  • 4
    $\begingroup$ $$I(b)=\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x}dx\iff I'(b)=\int_{-\frac\pi2}^{\frac\pi2}\frac{dx}{1+b\sin x}$$ $\endgroup$
    – Lucian
    Mar 27, 2014 at 9:16
  • $\begingroup$ I tried this but the new integral in b looked scary. $\endgroup$
    – lakshayg
    Mar 27, 2014 at 9:20

3 Answers 3


From the Taylor series expansion of $\ln(1+x)$ we know that $$\ln(1+b\sin x) = b\sin x - \frac{b^2\sin^2x}{2} +\frac{b^3\sin^3x}{3} - \cdots$$

Therefore $$\frac{\ln(1+b\sin x)}{\sin x} = b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots$$

Integrating both sides of the equation $$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \int^{\pi/2}_{-\pi/2} b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots \mathrm{d}x$$

Observe that $\int^{\pi/2}_{-\pi/2}\sin^nx\mathrm{d}x = 0$ if n is an odd number. The above equation is then equivalent to $$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \int^{\pi/2}_{-\pi/2} b +\frac{b^3\sin^2 x }{3}+\frac{b^5\sin^4 x }{5}\cdots \mathrm{d}x$$

As $\sin^{2n}x$ is an even function therefore $$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x =2 \int^{\pi/2}_0 b +\frac{b^3\sin^2 x }{3}+\frac{b^5\sin^4 x }{5}\cdots \mathrm{d}x$$

Using the fact that $\int_0^{\pi/2}\sin^nx\mathrm{d}x = \frac{(n-1)\cdot(n-3)\cdots3\cdot1}{n\cdot(n-2)\cdots4\cdot2}$ if n is even (See this page for the proof)

$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x =2\cdot\frac{\pi}{2}\Bigg[ b+\frac{b^3}{3}\frac{1}{2}+\frac{b^5}{5}\frac{3\cdot1}{4\cdot2}\cdots\Bigg]$$ $$=\pi\Bigg[ b+\frac{b^3}{3}\frac{1}{2}+\frac{b^5}{5}\frac{3\cdot1}{4\cdot2}\cdots\Bigg]$$ The term in the parenthesis is precisely the Taylor series expansion (Thanks to your edit. I was completely clueless at this point) for $\sin^{-1} b $. Therefore the integral is $$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \pi\sin^{-1} b $$

  • $\begingroup$ Knowing the answer this looks like back working. $\endgroup$
    – evil999man
    Mar 27, 2014 at 9:04
  • $\begingroup$ Actually it in not. I checked it only after your edit that it was indeed the expansion for arcsin(b). Anyway, thanks a lot. $\endgroup$
    – lakshayg
    Mar 27, 2014 at 9:11
  • $\begingroup$ @awesome. I am curious to see if you get a better answer. $\endgroup$ Mar 27, 2014 at 9:12
  • $\begingroup$ We have to see how the hell someone made this question? $\endgroup$
    – evil999man
    Mar 27, 2014 at 9:13
  • $\begingroup$ @LakshayGarg. Thanks for this answer. I also started using Taylor expansions but they were based on $x$ and not $\sin(x)$ (you can see how stupid I can be or I am). Cheers. $\endgroup$ Mar 27, 2014 at 9:14

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

With Weierstrass Substitution $\ds{t \equiv \tan\pars{x \over 2}}$: \begin{align} {\cal I}'\pars{b}&=\int_{-\pi/2}^{\pi/2}{\dd x \over 1 + b\sin\pars{x}} =\int_{-1}^{1}{2\,\dd t/\pars{1 + t^{2}} \over 1 + b\bracks{2t/\pars{1 + t^{2}}}} =2\int_{-1}^{1}{\dd t \over t^{2} + 2bt + 1} \\[3mm]&=2\int_{-1}^{1}{\dd t \over \pars{t + b}^{2} + 1 - b^{2}} =2\int_{-1 + b}^{1 + b}{\dd t \over t^{2} + 1 - b^{2}} \\[3mm]&={2 \over \root{1 - b^{2}}} \int_{\pars{-1 + b}/\root{1 - b^{2}}}^{\pars{1 + b}/\root{1 - b^{2}}} {\dd t \over t^{2} + 1} \\[3mm]&={2 \over \root{1 - b^{2}}}\bracks{% \arctan\pars{1 + b \over \root{1 - b^{2}}} -\arctan\pars{-1 + b \over \root{1 - b^{2}}}} \\[3mm]&={2 \over \root{1 - b^{2}}}\bracks{% {\pi \over 2} - \arctan\pars{\root{1 - b^{2}} \over 1 + b} -\arctan\pars{-1 + b \over \root{1 - b^{2}}}} \\[3mm]&={2 \over \root{1 - b^{2}}}\braces{% {\pi \over 2} - \overbrace{% \bracks{\arctan\pars{1 - b \over \root{1 - b^{2}}} +\arctan\pars{-1 + b \over \root{1 - b^{2}}}}}^{\ds{=\ 0}}} \\[3mm]&={\pi \over \root{1 -b^{2}}} \end{align}

Then $$\color{#00f}{\large% \int_{-\pi/2}^{\pi/2}{\ln\pars{1 + b\sin\pars{x}} \over \sin\pars{x}}\,\dd x =\pi\,\arcsin\pars{b}} $$


I have been trying to compute the antiderivative; I tried several change of variable with absolutely nos success; as already mentioned, integration by parts makes the problem still more complex.

Using a CAS (brut force), the antiderivative which came out is just a nightmare for me $$-\text{Li}_2\left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{b^2-1}-b}\right)-\text{Li}_2\left(-\frac{\tan \left(\frac{x}{2}\right)}{b+\sqrt{b^2-1}}\right)+\log \left(\tan \left(\frac{x}{2}\right)\right) \left(-\log \left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{b^2-1}+b}+1\right)-\log \left(\left(\sqrt{b^2-1}+b\right) \tan \left(\frac{x}{2}\right)+1\right)+\log (b \sin (x)+1)+\log \left(1-i \tan \left(\frac{x}{2}\right)\right)+\log \left(1+i \tan \left(\frac{x}{2}\right)\right)\right)+\frac{1}{2} \text{Li}_2\left(-\tan ^2\left(\frac{x}{2}\right)\right)$$ Computing the integral leads, after simplifications, to $$-\text{Li}_2\left(\frac{1}{\sqrt{b^2-1}-b}\right)-\frac{1}{2} \text{Li}_2\left(\frac{1}{\left(b+\sqrt{b^2-1}\right)^2}\right)+2 \text{Li}_2\left(\frac{1}{b+\sqrt{b^2-1}}\right)+\text{Li}_2\left(b+\sqrt{b^2-1}\right) $$ which is a real valued function of $b$. Trying to use the definition and properties of the dilogarithm took me to a dead end.

Thanks to your edit and clever remark, I verified that the result is exactly what you said, that is to say that $$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx = \pi \sin ^{-1}(b)$$

I checked that the Taylor series built at $b=0$ are identical.

Congratulations for having found the result. I am really sorry to have not been able to simplify these monsters.

Added later after Lucian's suggestion

After Lucian's so good suggestion, let us consider $$I'(b)=\int_{-\frac\pi2}^{\frac\pi2}\frac{dx}{1+b\sin x}$$ First of all, the antiderivative is given by $$\frac{2 \tan ^{-1}\left(\frac{b+\tan \left(\frac{x}{2}\right)}{\sqrt{1-b^2}}\right)}{\sqrt{1-b^2}}$$ Integrated between the given bounds $$I'(b)=\frac{2 \left(\tan ^{-1}\left(\sqrt{\frac{1-b}{1+b}}\right)+\tan ^{-1}\left(\sqrt{\frac{1+b}{1-b}}\right)\right)}{\sqrt{1-b^2}}=\frac{\pi }{\sqrt{1-b^2}}$$ and the final result for $I(b)$.

  • $\begingroup$ This should be not be that tough... $\endgroup$
    – evil999man
    Mar 27, 2014 at 9:02
  • $\begingroup$ @Awesome. I agree with you. Fortunately, you receive a nice answer from Lakshay Garg. $\endgroup$ Mar 27, 2014 at 9:10
  • $\begingroup$ @Claude. And my search for a better solution comes to an end. Well done. $\endgroup$
    – lakshayg
    Mar 27, 2014 at 11:50

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