Integral $\int_0^b \frac{1-\cos ax}{x} dx$

Question

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.

Are you supposed to know about the cosine integral function ? — Claude Leibovici, Aug 22 '19 at 07:33
Try using Feynman trick : differentiate with respect to $a$. It could help (may be). — Claude Leibovici, Aug 22 '19 at 07:37
This integral has no closed form and the answer you obtained is the most general. — Mostafa Ayaz, Aug 22 '19 at 07:45
One parameter between $a$ and $b$ is redundant (it can be removed through a substitution), hence you are asking for a primitive of $\frac{1-\cos x}{x}$, that is the same as asking for a primitive of $\frac{\sin x}{x^2}$. — Jack D'Aurizio, Aug 22 '19 at 11:21

Infiniticism · Accepted Answer · 2020-11-12T17:00:56.380

2

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.

edited Nov 12 '20 at 17:00

answered Sep 15 '19 at 11:33

Infiniticism

1 Answers1