# Evaluation of $\int_0^\pi \! \ln\left(1-2\alpha\cos x+\alpha^2\right) \mathrm{d}x$

I have got a trouble with integral $$\int_0^\pi \! \ln\left(1-2\alpha\cos x+\alpha^2\right) \, \mathrm{d}x,\quad |\alpha|<1.$$

My teacher said there are two ways of solving such ones, if there is no straight solution. The first one is when you simply differentiate by $\alpha$, then, if it helps, integrate by $x$ and then again integrate by $\alpha$. The second is more sophisticated, then you substitute any allowed $\alpha$ and integrate, but I did not get it properly. Can anyone help me with this one? Thank you in advance.

• Is the parameter $a$ assumed to be a real number? – user64494 Oct 28 '14 at 19:18
• Yes, it is a real number. – Dankevich Oct 28 '14 at 19:19
• Maple outputs $0$ for $a < 1, a > 0$ and $$\pi \,\ln \left( -a \right) +\ln \left( -{a}^{-1} \right) \pi$$ for $a < 0, a > -1.$ The latter equals $0$ too. – user64494 Oct 28 '14 at 19:24
• See en.wikipedia.org/wiki/Differentiation_under_the_integral_sign This is solved example number 3 – user_of_math Oct 28 '14 at 19:25

Consider integrating

$$\,I(\alpha)=\int_0^\pi\,\ln(1-2\alpha\cos x+\alpha^2)\;dx \qquad |\alpha| > 1.$$

Now,

\begin{align} \frac{d}{d\alpha}\,I(\alpha) &=\int_0^\pi \frac{-2\cos x+2\alpha }{1-2\alpha \cos x+\alpha^2}\;dx\, \\[8pt] &=\frac{1}{\alpha}\int_0^\pi\,\left(1-\frac{1-\alpha^2}{1-2\alpha \cos x+\alpha^2}\,\right)\,dx \\[8pt] &=\frac{\pi}{\alpha}-\frac{2}{\alpha}\left[\,\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\right]\,\bigg|_0^\pi. \end{align}

As $x$ varies from $0$ to $\pi$, $\left[\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right]\,$ varies through positive values from $0$ to $\infty$ when $|\alpha| < 1$ and $\left[\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right]\,$ varies through negative values from $0$ to $-\infty$ when $|\alpha| > 1$.

Hence,

$$\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=\frac{\pi}{2}\,$$ when $|\alpha| < 1$. $$\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=-\frac{\pi}{2}\,$$ when $|\alpha| > 1$.

Therefore,

$$\frac{d}{d\alpha}\,I(\alpha)\,=0\,$$ when $|\alpha| < 1$ and $$\frac{d}{d\alpha}\,I(\alpha)\,=\frac{2\pi}{\alpha}\,$$ when $|\alpha| > 1$.

Upon integrating both sides with respect to $\alpha$, we get $I(\alpha) = C_1$ when $|\alpha| < 1$ and $I(\alpha) = 2\pi \ln|\alpha| + C_2$ when $|\alpha| > 1$.

$C_1$ may be determined by setting $\alpha=0$ in $I(\alpha)$:

$$I(0) =\int_0^\pi \ln(1)\;dx =\int_0^\pi 0\;dx=0$$

Thus, $C_1=0$. Hence, $I(\alpha)=0$ when $|\alpha| < 1$.

To determine $C_2$ in the same manner, we should need to substitute in a value of $\alpha>1$ in $I(\alpha)$. This is somewhat inconvenient. Instead, we substitute $\alpha = \frac{1}{\beta}$, where $|\beta| < 1$. Then,

\begin{align} I(\alpha) &=\int_0^\pi\left(\ln(1-2\beta \cos x+\beta^2)-2\ln|\beta|\right)\;dx\ \\[8pt] &=0-2\pi\ln|\beta|\, \\[8pt] &=2\pi\ln|\alpha|\, \end{align}

Therefore, $C_2 = 0$ and $I(\alpha)=2\pi\ln|\alpha|$ when $|\alpha| > 1$.

$$I(\alpha)=\begin{cases} 0 &,\quad|\alpha| < 1\\[10pt] 2\pi\ln|\alpha| &,\quad |\alpha| > 1 \end{cases}$$

The foregoing discussion, of course, does not apply when $\alpha=\pm1$, since the conditions for differentiability are not met.

Source : Wikipedia