1But how does this help me in question ?
Prove the following (a2<1)
a)\int _{0} ^ { \pi } ln(1-2a \cos \theta +a^2) d \theta=\begin{Bmatrix} 0 & a<1\\ 2\pi lna & a>1 \end{Bmatrix}
b)\int _{0} ^ { \pi } ln(1-2a \cos \theta +a^2)\cos n \theta d \theta= - \frac {\pi a^n}{n}
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4 Answers
62I(a)=\int _{0} ^ { \pi } ln(1-2a \cos \theta +a^2) d \theta\\ I=\int _{0} ^ { \pi } ln(1-2a \cos (\pi-\theta) +a^2) d \theta\\ I=\int _{0} ^ { \pi } ln(1+2a \cos \theta +a^2) d \theta\\ 2I(a)=\int _{0} ^ { \pi } ln((1+a^2)^2-4a^2 \cos^2 (\theta)) d \theta \text{ Adding lines 1 and 3}\\ 2I(a)=\int _{0} ^ { \pi } ln(1+a^4+2a^2-4a^2 \cos^2 (\theta)) d \theta \\ 2I(a)=\int _{0} ^ { \pi } ln\left( 1+a^4+2a^2(1-2 \cos^2 \theta)\right) d \theta \\ 2I(a)=\int _{0} ^ { \pi } ln\left( 1+a^4-2a^2 \cos 2\theta\right) d \theta \\
\\2I(a)=\int _{0} ^ { \pi } ln\left( 1+a^4-2a^2 \cos 2\theta)\right) d \theta \\ 2I(a)=\int _{0} ^ { 2\pi } ln\left( 1+a^4-2a^2 \cos t)\right) dt/2 \\ 2I(a)=2\int _{0} ^ { \pi } ln\left( 1+a^4-2a^2 \cos t)\right) dt/2 \\ 2I(a)=I(a^2)
341Well that is all you need. I will use Nishant sir's notation and the functional relationship derived by him
First note that I(a) = I(-a) (prove it!). So we can restrict ourselves to a>0.
Now, consider the case 0<a<1
We have I(a) = \frac{I(a^2)}{2} = \frac{I(a^4)}{4} = ....
The numerator tends to I(0) which is 0. and the denominator becomes arbitrarily large. Hence I(a) can be made arbitrarily close to zero. In other words I(a) = 0 for 0<a<1.
If a>1, we write the integrand as \ln (a^2-2a \cos \theta + 1) = 2 \ln a + \ln \left(\frac{1}{a^2} - 2 \frac{\cos \theta}{a} + 1 \right) = 2 \ln b + \ln \left(b^2 - 2b \cos \theta + 1 \right) where 0<b<1 (sorry 2 ln a not 2 ln b)
Now, the second integrand reduces to zero, the first integrand becomes 2 \pi \ln a
