Let $f$ be the function from $\mathbb { R }$ to $\mathbb { R }$ defined by $f ( x ) = \int _ { 0 } ^ { \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta$. Deduce that $$f ( x ) = \begin{cases} 2 \pi \ln ( | x | ) & \text { if } | x | > 1 \\ 0 & \text { if } | x | < 1 \end{cases}$$ One will first determine coefficients $A$ and $B$ as functions of $x$ such that $\frac { ( x + 1 ) T + ( x - 1 ) } { \left( ( x + 1 ) ^ { 2 } T + ( x - 1 ) ^ { 2 } \right) ( T + 1 ) } = \frac { A } { ( x + 1 ) ^ { 2 } T + ( x - 1 ) ^ { 2 } } + \frac { B } { T + 1 }$ for all $T \in \mathbb { R }$ such that these fractions are defined.
Let $f$ be the function from $\mathbb { R }$ to $\mathbb { R }$ defined by $f ( x ) = \int _ { 0 } ^ { \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta$.
Deduce that
$$f ( x ) = \begin{cases} 2 \pi \ln ( | x | ) & \text { if } | x | > 1 \\ 0 & \text { if } | x | < 1 \end{cases}$$
One will first determine coefficients $A$ and $B$ as functions of $x$ such that $\frac { ( x + 1 ) T + ( x - 1 ) } { \left( ( x + 1 ) ^ { 2 } T + ( x - 1 ) ^ { 2 } \right) ( T + 1 ) } = \frac { A } { ( x + 1 ) ^ { 2 } T + ( x - 1 ) ^ { 2 } } + \frac { B } { T + 1 }$ for all $T \in \mathbb { R }$ such that these fractions are defined.