Let $a$, $b$ and $c$ be positive real scalars such that $a>b,c$ and consider the following integral $$ \int_{t_0}^{t_1} \log\left(\frac{a-c}{a-b\sin(\tau)}\right)\,\mathrm{d}\tau. $$

My question. Does there exist some sort of nice closed-form expression of the above integral? If so, does there also exist a "nice" closed-form expression of the following multiple integral $$ \int_{t_0}^{t_1} \int_{\tau_{1,0}}^{\tau_1} \cdots \int_{\tau_{k-1,0}}^{\tau_{k-1}} \log\left(\frac{a-c}{a-b\sin(\tau_k)}\right)\,\mathrm{d}\tau_k \cdots\,\mathrm{d}\tau_2\,\mathrm{d}\tau_1 \ \ \ ? $$

  • $\begingroup$ Doing a bit of testing by plugging the indefinite integral into Wolfram Alpha, it would appear that this function is integrable, and closed-form. It's just not very nice. Here's what I plugged in: wolframalpha.com/input/?i=integrate+ln((a-c)%2F(a-b*sin(x)))+dx Unfortunately it does not restrict to your specific case of $a\gt b,c$, but the last form expressed on the page does at least restrict $a,b,c\gt0$. $\endgroup$ – Ispil May 11 '18 at 2:26
  • $\begingroup$ @Ispil: I've tried with Wolfram Alpha as well before posting this question. The resulting expression seems quite messy, that is why I hope that a "nicer" expression exists. $\endgroup$ – Ludwig May 11 '18 at 2:39

Considering $$I=\int \log\left(\frac{a-c}{a-b\sin(\tau)}\right)\,d\tau$$ use the usual tangent half-angle substitution to get $$\frac 12I =\int \frac{\log \left( (a-c)\left(t^2+1\right)\right)}{t^2+1}\,dt-\int\frac{\log \left(a t^2-2 b t+a\right)}{t^2+1}\,dt$$

Looking at @Maxim's answer to this question,and quoting

"Factoring the polynomials and formally expanding the logarithm reduces the indefinite integral to a sum of integrals of the form" $$\int \frac {\log (u)} {u + a}\, du = \log (u)\, \log \left( 1 + \frac u a \right) + \operatorname{Li}_2 \left( - \frac u a \right)$$


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