Hey, how are you? I was troubled by calculating $\displaystyle f(a)=\int_0^\pi \log(1+a \cos x)dx.$ Here, $\displaystyle |a|<1$ is real.
I do know we can differentiating under the integral, however, I could not see any more...
Hey, how are you? I was troubled by calculating $\displaystyle f(a)=\int_0^\pi \log(1+a \cos x)dx.$ Here, $\displaystyle |a|<1$ is real.
I do know we can differentiating under the integral, however, I could not see any more...
That seems like a reasonable approach. Differentiate with respect to a and exchange the derivative and integral. You should get a rational function of cos x which you can integrate. Then you will have to integrate with respect to a and evaluate the constant by looking at a=0.
So what do you get for the derivative of log(1+a cos x) with respect to a?
- Hollywood
Yes, I think that would qualify as a difficult integral. But it's not impossible. You can substitute $\displaystyle u=\tan\frac{x}{2}$ to convert any rational combination of trig functions to a rational function of u, which can then be integrated using partial fractions. But then it might not be possible to integrate the result with respect to a.
I like your method.
- Hollywood