cosine integral

Question

Show that $$\int_0^x \frac{1-\cos(t)}{t}=\gamma+\ln(x)-\operatorname{Ci}(x)$$ where $$\operatorname{Ci}(x)=-\int_x^\infty \frac{\cos(t)}{t} \, dt$$ and gamma is an euler-mascheroni constant. I did as follows: $$\int_0^x \frac{1-\cos(t)}{t} \, dt=\int_0^x \frac{1}{t} \,dt-\int_0^x \frac{\cos(t)}{t}\,dt$$ $$=\int_0^x \frac{1}{t}\,dt-\int_0^\infty \frac{\cos(t)}{t}\,dt+\int_x^\infty \frac{\cos(t)}{t} \, dt=\int_0^x \frac{1}{t}-\int_0^\infty \frac{\cos(t)}{t}-\operatorname{Ci}(x)$$

I am stuck with the last two integrals. How do I proceed? A shorthand notation tells me that I get $\ln(x)-\ln(0)-\Gamma(0)$ which is wrong since $\ln(0)$ and $\Gamma(0)$ are undefined unless $$\ln(0)+\Gamma(0)=-\gamma$$

Answer 1

Here is an approach.

$$ \int_0^x \frac{1-\cos(t)}{t}dt = \sum_{k=1}^{\infty}\frac{(-1)^k}{(2 k)!}\int_0^x t^{2k-1}\,dt = \dots\,. $$

Note: The Taylor series of $\cos(t)$ is

$$ \cos(t) = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k)!}t^{2k}. $$

Answer 2

You can't break up the interval from $0 $ to $x$ like that because the integral of $1/x$ diverges, but you can try $\epsilon$ to $x$ for $\epsilon > 0$.

$$ \int_\epsilon^x \dfrac{1-\cos(t)}{t}\ dt + {\rm Ci}(x) = \ln(x) - \ln(\epsilon) - \int_{\epsilon}^\infty \dfrac{\cos(t)}{t}\ dt = \ln(x) - \ln(\epsilon) + {\rm Ci}(\epsilon)$$

So now what you need is to show that $$\lim_{\epsilon \to 0+} \left({\rm Ci(\epsilon)} - \ln(\epsilon)\right) = \gamma$$