calculus - Evaluation of $int_0^pi ! lnleft(1-2alphacos x+alpha^2right) mathrm{d}x$

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.

Answer

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.

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Source : Wikipedia