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

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 $$\lim_{\epsilon \to 0+} \left({\rm Ci(\epsilon)} - \ln(\epsilon)\right) = \gamma$$

Answer 2

Let $\displaystyle f(z)=\frac{e^{iz}}{z}$, $z\in \mathbb{C}$. Then, for $R>\varepsilon>0$, Cauchy's Integral Theorem guarantees that

$$\begin{align} 0&=\int_\varepsilon^R \frac{e^{ix}}{x}\,dx +\int_0^{\pi/2}\frac{e^{iRe^{i\phi}}}{Re^{i\phi}}\,iRe^{i\phi}\,d\phi+\int_R^\varepsilon \frac{e^{-x}}{ix}\,i\,dx+\int_{\pi/2}^0 \frac{e^{i\varepsilon e^{i\phi}}}{\varepsilon e^{i\phi}}\,i\varepsilon e^{i\phi}\,d\phi\\\\ &=\int_\varepsilon^R \frac{e^{ix}}{x}\,dx -\int_\varepsilon^R \frac{e^{-x}}{x}\,dx-i\frac\pi2+O(\varepsilon)+O\left(\frac1R\right)\tag1 \end{align}$$

Taking the real part of both sides of $(1)$ and integrating by parts the integral $\int_\varepsilon^R \frac{e^{-x}}{x}\,dx$ with $u=e^{-x}$ and $v=\log(x)$, we find

$$\begin{align}-\int_x^R \frac{\cos(x')}{x'}\,dx'&=\log(x)-\int_\varepsilon^R e^{-x}\log(x)\,dx+\int_\varepsilon^x\frac{\cos(x')-1}{x'}\,dx'\\\\ &-\log(\varepsilon) -e^{-R}\log(R)+e^{-\varepsilon}\log(\varepsilon)+O\left(\frac1R\right)+O(\varepsilon)\tag2 \end{align}$$

Letting $R\to\infty$ and $\varepsilon\to 0$ in $(2)$ yields

$$\text{Ci}(x)=\log(x)+\gamma+\int_0^x \frac{\cos(x')-1}{x'}\,dx'\tag3$$

whereupon rearranging $(3)$ yields the coveted relationship.

---

Alternatively, we can use either the Laplace Transform or "Feynman's Trick" to show that

$$\int_0^\infty \frac{\cos(x)-e^{-x}}{x}\,dx=\int_0^\infty \left(\frac{x}{x^2+1}-\frac1{x+1}\right)=0\tag4$$

Starting with $(4)$, is tantamount to starting with the real part of $(1)$ and we are done.

Answer 3

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

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