Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me to solve this integral: $$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$

I managed to calculate an indefinite integral of the left part: $$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$ where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.

I tried a numerical integration and it looks plausible that $$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$ but I have no idea how to prove it.

share|improve this question
2  
Mathematica cannot find a closed form for this integral. The last conjectural equality is correct up to at least 750 decimal digits. –  Vladimir Reshetnikov May 14 '13 at 3:37

3 Answers 3

up vote 9 down vote accepted

Let $$y=\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x},$$ then, solving this with respect to $x$, we get $$x=\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}.$$ So, $$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx=\int_0^\infty\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\mathrm dy.$$ The latter integral can be solved by Mathematica and yields $$\pi\log54.$$


Of course, we want to prove that the result returned by Mathematica is correct.

The following statement is provably true, that can be checked directly by taking derivatives of both sides: $$\int\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\mathrm dy =\\ \frac{1}{2} i \left(2 \text{Li}_2\left(\frac{iy}{8}+\frac{1}{2}\right)+\text{Li}_2\left(\frac{iy}{6}+\frac{1}{3}\right)+2\text{Li}_2\left(\frac{iy}{6}+\frac{2}{3}\right)+\text{Li}_2\left(\frac{iy}{4}+\frac{1}{2}\right)+\text{Li}_2\left(\frac{2i}{y-2 i}\right)-\text{Li}_2\left(-\frac{2 i}{y+2i}\right)-\text{Li}_2\left(-\frac{1}{6} i (y+2i)\right)-\text{Li}_2\left(-\frac{1}{4} i (y+2i)\right)-2 \left(-\text{Li}_2\left(-\frac{2i}{y-4 i}\right)+\text{Li}_2\left(\frac{2 i}{y+4i}\right)+\text{Li}_2\left(-\frac{1}{8} i (y+4i)\right)+\text{Li}_2\left(-\frac{1}{6} i (y+4i)\right)\right)\right)+\pi \left(\frac{1}{2}\log \left(3 \left(y^2+4\right)\right)+\log\left(\frac{3}{64}\left(y^2+16\right)\right)\right)+\log \left(4\left(y^2+4\right)\right) \arctan\left(\frac{y}{4}\right)-\left(\log576-2\log \left(y^2+16\right)\right) \arctan\left(\frac{4}{y}\right)+\log\left(y^2+4\right) \text{arccot}\left(\frac{6y}{8-y^2}\right)-\arctan\left(\frac{2}{y}\right)\log12 +\arctan\left(\frac{y}{2}\right)\log2$$

The remaining part is to calculate $\lim\limits_{y\to0}$ and $\lim\limits_{y\to\infty}$ of this expression, which I haven't done manually yet, but it looks like a doable task.

share|improve this answer
    
But how to solve the latter without the Mathematica? –  Martin Gales May 14 '13 at 10:54
    
Good question! I am still working on it, and going to publish the solution once I'm done. –  Vladimir Reshetnikov May 14 '13 at 16:51

Here's one way to go.

First, note that $$\begin{eqnarray*} \int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx &=& \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx +\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx. \end{eqnarray*}$$ For now I'll simply claim that \begin{equation*} \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx = \pi\log 64.\tag{1} \end{equation*} (I would be surprised if this integral has not been handled somewhere on this site.) But $$\begin{eqnarray*} \int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx &=& \int_0^\pi\frac{3x}{\sin x} \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k}} \sin^{2k}x \ \mathrm dx \\ &=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}} \int_0^\pi x \sin^{2k-1}x \ \mathrm dx \\ &=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}} \frac{\pi^{3/2}\Gamma(k)}{2\Gamma(k+1/2)} \\ &=& -\pi \sum_{k=1}^\infty \frac{1}{3^{2k-1}2k(2k-1)} \\ &=& -\pi \log \frac{32}{27}. \end{eqnarray*}$$ (The last sum can be found by standard methods. Schematically, $\sum \frac{a^{2k-1}}{2k(2k-1)} = \sum \int {\mathrm da} \frac{a^{2k-2}}{2k}$.) Thus, the integral is $\pi \log 54$ as claimed.


Proof of (1): We have $$\begin{eqnarray*} \int_0^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx &=& \int_{0^+}^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx \\ &=& 3\int_{0^+}^\pi x \cot\frac{x}{2} \ \mathrm dx \hspace{5ex}\textrm{(double angle formulas)} \\ &=& 12 \int_{0^+}^{\pi/2} t\cot t \ \mathrm dt \hspace{5ex} (t = x/2) \\ &=& -12\int_{0^+}^{\pi/2} \log\sin t \ \mathrm dt \hspace{5ex}\textrm{(integrate by parts)} \\ &=& -6\int_{0^+}^{\pi/2} \log\sin^2 t \ \mathrm dt \\ &=& -6\int_{0^+}^{\pi/2} \log(1-\cos^2 t) \ \mathrm dt \\ &=& 6 \int_{0^+}^{\pi/2} \sum_{k=1}^\infty \frac{1}{k}\cos^{2k}t \ \mathrm dt \hspace{5ex}\textrm{(series for log)} \\ &=& 6\sum_{k=1}^\infty \frac{1}{k} \int_{0^+}^{\pi/2} \cos^{2k}t \ \mathrm dt \hspace{5ex} \textrm{(Tonelli's theorem)}\\ &=& 6\sum_{k=1}^\infty \frac{1}{k} \frac{\sqrt{\pi}\Gamma(k+1/2)}{2\Gamma(k+1)} \\ &=& 3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1}\frac{2k-1}{k} \\ &=& \pi \log 64. \end{eqnarray*}$$ Note that $$\begin{eqnarray*} 6\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} &=& -6\pi \left[\sum_{k=0}^\infty {1/2 \choose k}(-1)^{k} - 1\right] \\ &=& -6\pi[(1-1)^{1/2} - 1] \\ &=& 6\pi \end{eqnarray*}$$ and $$\begin{eqnarray*} -3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} \frac{1}{k} &=& 3\pi \sum_{k=1}^\infty {1/2\choose k}(-1)^k \int_0^1 x^{k-1} \ \mathrm dx \\ &=& 3\pi \int_0^1 \frac{1}{x} \left[ \sum_{k=0}^\infty {1/2\choose k}(-1)^k x^{k} -1 \right] \ \mathrm dx \\ &=& 3\pi \int_0^1 \frac{1}{x} \left( \sqrt{1-x} -1 \right) \ \mathrm dx \\ &=& 3\pi(-2+\log 4) \\ &=& -6\pi + \pi\log 64. \end{eqnarray*}$$

share|improve this answer

Begin with $u$-substitution using $$\begin{align} u & =x\\ dv & = \frac{3\cos(x)+\sqrt{8+\cos^2(x)}}{\sin(x)}\,dx\end{align}$$ so that $du=dx$, and my CAS tells me (which I suppose could be verified through differentiation and identities) that $$\begin{align} v & = \sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x)) \end{align}$$

Now we have $$\begin{align} \left[x\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)\right]_0^\pi\\ -\int_0^\pi\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)dx \end{align}$$

and most of the integral part can be evaluated by taking advantage of symmetry about $\pi/2$:

$$\begin{align} \left[x\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)\right]_0^\pi\\ -3\int_0^\pi\ln(1-\cos(x))dx \end{align}$$

($\sinh^{-1}$ is odd and $\cos(x)$ has odd symmetry about $\pi/2$. For the logarithmic term, the input to $\ln()$ at $x$ is the reciprocal of the input at $\pi/2-x$.)

Some of the nonintegral-part can be cleanly evaluated:

$$\begin{align} \pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\left[3x\ln(1-\cos(x))\right]_0^\pi\\ -3\int_0^\pi\ln(1-\cos(x))dx \end{align}$$

and now moving the "unclean" part back into an integral: $$\begin{align} \pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\int_0^\pi\left(3\ln(1-\cos(x))+\frac{3x\sin(x)}{1-\cos(x)}\right)\,dx\\ -3\int_0^\pi\ln(1-\cos(x))dx\\ =\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\int_0^\pi\frac{3x\sin(x)}{1-\cos(x)}\,dx \end{align}$$

My CAS says this is

$$\begin{align} \pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\pi\ln(64) \end{align}$$

which is the only thing the CAS does that I don't quite get. But it's nothing special about endpoints: even WA can give an antiderivative if we can use the dilogarithm. It looks like an integral that might even appear somewhere on this site. A conversion of the arcsinh and logarithm rules yields

$$\begin{align} \pi\ln(2^{-\frac{1}{2}})+\pi\ln\left(\frac{27}{8^{3/2}}\right)+\pi\ln(64)=\pi\ln(54) \end{align}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.