Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$

Question

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.

Answer 1

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.

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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 $$\begin{eqnarray*} -3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} \frac{1}{k} &=& 3\pi \sum_{k=1}^\infty {1/2\choose k}(-1)^k \int_0^1 x^{k-1} \ \mathrm dx \\ &=& 3\pi \int_0^1 \frac{1}{x} \left[ \sum_{k=0}^\infty {1/2\choose k}(-1)^k x^{k} -1 \right] \ \mathrm dx \\ &=& 3\pi \int_0^1 \frac{1}{x} \left( \sqrt{1-x} -1 \right) \ \mathrm dx \\ &=& 3\pi(-2+\log 4) \\ &=& -6\pi + \pi\log 64. \end{eqnarray*}$$

Answer 2

Begin with integration by parts 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

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

Answer 3

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 $$\pi\log54.$$

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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.

Answer 4

Substitute $t=\cos x$ \begin{align} &\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x \ dx =\int_{-1}^1 \cos^{-1}t\ \frac{3t+\sqrt{8+t^2}}{1-t^2}dt\\ =& \int_{-1}^1 \cos^{-1}t\ d\bigg[3\tanh^{-1}\frac t{3\sqrt{t^2+8}+8}-\sinh^{-1}\frac t{\sqrt8}-3\ln(1-t) \bigg]\\ \overset{ibp}=&\ \pi\ln\frac{27}4 -\int_{-1}^1\frac{3\ln(1-t)}{\sqrt{1-t^2}}dt= \pi\ln\frac{27}4 - (-\pi\ln 8)=\pi\ln54 \end{align}

More generally, for $b>1$

\begin{align} &\int_0^\pi\frac{b\cos x+\sqrt{b^2-\sin^2 x}}{\sin x}x\ dx=\pi\bigg( b\ln\frac{4b}{\sqrt{b^2-1}}-\ln\frac{b+1}{ \sqrt{b^2-1}}\bigg) \end{align} and the integral in question is a special case with $b=3$.

Answer 5

$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx=\int_0^\pi \color{red}{x}\color{blue}{\frac{3\cos x+\sqrt{8+\cos^2 x}}{1-\cos^2 x}\sin x\,dx}$$ Integration by parts with $u=x$ and $dv$ the blue expression: $$v=\frac32\ln\left(\frac{3\sqrt{\cos^2x+8}-\cos x+8}{3\sqrt{\cos^2x+8}+\cos x+8}\right)+3\ln(1-\cos x)+\sinh^{-1}\left(\frac{\cos x}{2\sqrt 2}\right).$$ Then $uv\vert_0^{\pi}=\pi(\ln27-\ln4)$. And in $\int_0^{\pi}dv$ only $3\ln(1-\cos x)$ term contributes, giving $-\pi\ln8$. Hence, total is $$\pi(\ln27-\ln4+\ln8)=\pi\ln54.$$

Answer 6

$$I=\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ dx=\int_0^\pi x(3\cot x+\sqrt{8\csc^2x +\cot^2x})dx$$ Substitute $x → \pi-x$ and add two versions of I $$2I=\int_0^\pi(3x\cot x+x\sqrt{8\csc^2x +\cot^2x})dx+\int_0^\pi(3(x-\pi)\cot x+(\pi-x\sqrt{8\csc^2x +\cot^2x}))dx$$ $$=\int_0^\pi(-6(\pi/2-x)\cot x+\pi\sqrt{8\csc^2x +\cot^2x})dx$$ Substitute $x → \pi/2-x$ and use even-odd properties $$I=\int_0^{\pi/2}(-6x\tan x+\pi\sqrt{8+9\tan^2x})dx$$ Substitute $u=\cot x$ $$I=\pi\int_0^\infty\frac{\sqrt{9+8u^2}-3}{u(1+u^2)}du+6\int_0^\infty\frac{\arctan u}{u(1+u^2)}du$$ The first integral can be easily solved with $v=\sqrt{9+8u^2}-3$ and the second integral is solved here $$I=\pi\ln(27/4)+3\pi\ln(2)=\pi\ln(54)$$

Answer 7

Since @Vladimir_Reshetnikov gave an approach by transforming the integral into $$\int_0^{\infty}\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\, dy$$ I kinda don't want to waste it. Here's the solution to this integral.

Notice that $\text{arctan}(\frac{6y}{8-y^2})=\arctan(y/2)+\arctan(y/4)$

$$I=\int_0^{\infty}\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\, dy$$ $$=\int_0^{\infty}\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\pi-\text{arctan}\frac{6y}{8-y^2}\right)\, dy$$

$$=\int_0^{\infty}\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\pi-\arctan(y/2)-\arctan(y/4)\right)\, dy$$ $$=\int_0^{\infty}\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\arctan(2/y)+\arctan(4/y)\right)\, dy$$ substituting $1/y=x$ gives

$$I=\int_0^{\infty}\frac{6(8x^2+1)}{x(4x^2+1)(16x^2+1)}(\arctan(2x)+\arctan(4x))\, dx$$

Note the partial fraction decomposition $$\frac{6(8x^2+1)}{(4x^2+1)(16x^2+1)}=\frac{4}{16x^2+1}+\frac{2}{4x^2+1}$$ Which the right hand side is $\frac{d}{dx}(\tan^{-1}(2x)+\tan^{-1}(4x))$

$$I=\int_0^{\infty} \frac{1}{x}(\tan^{-1}(2x)+\tan^{-1}(4x))\frac{d}{dx}(\tan^{-1}(2x)+\tan^{-1}(4x))\, dx$$

Integration by parts gives

$$=\frac{1}{2}\int_0^{\infty} \frac{(\tan^{-1}(2x)+\tan^{-1}(4x))^2}{x^2}\, dx=\int_0^{\infty} \frac{(\tan^{-1}(x)+\tan^{-1}(2x))^2}{x^2}\, dx$$

$$I=3{\underbrace{\int_0^{\infty}\frac{\text{arctan}^2 x}{x^2}\, dx}_{J_1}}+2{\underbrace{\int_0^{\infty}\frac{\text{arctan}(x)\text{arctan}(2x)}{x^2}\, dx}_{J_2}}$$

Integrating by parts gives

$$J_1=\int_0^{\infty}\frac{\text{arctan}^2 x}{x^2}\, dx=2\int_0^{\infty}\frac{\tan^{-1} x}{x(1+x^2)}\, dx$$ $$=2\int_0^{\infty}\tan^{-1}(x)\left(\frac{1}{x}-\frac{x}{1+x^2}\right)\, dx$$

integrating by parts again gives

$$J_1=-\int_0^{\infty}\frac{2\log x-\log(1+x^2)}{1+x^2}\, dx=\int_0^{\infty}\frac{\log(1+x^2)}{1+x^2}\, dx$$

$$J_1=\pi\log(2)$$

We will need the following result to solve for $J_2$ \begin{equation}\label{a} \int_0^{\infty}\frac{\log(1+a^2 x^2)}{1+b^2 x^2}\, dx=\frac{\pi}{b}\log\left(1+\frac{a}{b}\right) \end{equation} Which is easily proved using Leibniz rule.

We use the same approach to compute $J_2$:

$$J_2=\int_0^{\infty}\frac{\text{arctan}(x)\text{arctan}(2x)}{x^2}\, dx$$ IBP gives $$=\int_0^{\infty}\frac{2\tan^{-1} x}{x(1+x^2)}\, dx+\int_0^{\infty}\frac{\tan^{-1}(2x)}{x(1+x^2)}\, dx$$ $$=\int_0^{\infty}\tan^{-1}(x)\left(\frac{2}{x}-\frac{x}{1+4x^2}\right)\, dx+\int_0^{\infty}\tan^{-1}(2x)\left(\frac{1}{x}-\frac{x}{1+x^2}\right)\, dx$$

IBP again gives

$$=-\frac{\pi}{2}\log(4)+\int_0^{\infty}\frac{\log(1+4x^2)}{1+x^2}\, dx-2\int_0^{\infty}\frac{\log x}{1+4x^2}\, dx+\int_0^{\infty}\frac{\log(1+x^2)}{1+4x^2}\, dx$$ $$=-\pi\log(2)+\pi\log 3+\frac{\pi}{2}\log 2+\frac{\pi}{2}\log\left(\frac{3}{2}\right)$$

$$J_2=-\pi\log(2)+\frac{3\pi}{2}\log(3)$$

Putting results together gives

$$I=3J_1+2J_2=3\pi\log 2-2\pi\log 2+3\pi \log 3$$

Which proves the conjectured result.

$$I=\pi\log 54$$