How can I evaluate this?
$$ I=\int_0^b \frac{1-\cos ax}{x} dx$$
I tried the following two approaches that do not work well:
-
Use the power series expansion
$$ \frac{1-\cos ax}{x}=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n)!} a^{2n}x^{2n-1}$$
and integrate term by term to obtain
$$I=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n(2n)!}(ab)^{2n}.$$
But how can I get a closed form expression? -
Use the fact that
$$ \frac 1x=\int_0^\infty e^{-tx} dt$$ and express $I$ as a double integral. After changing the order of integration, I obtained
$$ I=\int_0^\infty \frac{1-e^{-bt}}{t}-\frac{1}{t^2+a^2}\{e^{-bt}(a\sin ab-t\cos ab)+t\}dt$$
which is more compicated.
Best Answer
One have $\int_0^b \frac{1-\cos (a x)}{x} \, dx=-\text{Ci}(a b)+\log (a b)+\gamma$ where Ci is the cosine integral.