$\newcommand{\d}{\,\mathrm{d}}\newcommand{\cn}{\mathrm{Cin}}\newcommand{\ci}{\mathrm{Ci}}$I am very new to the Ci, Si functions – I saw these an hour ago! In particular, I saw an identity involving them as necessary in another proof that I was reading, but I am unsure how to prove this identity for myself. I made progress, but reached a difficult limit.
$$\begin{align}\cn(x)&=-\int_x^\infty\frac{\cos(t)}{t}\d t\\\ci(x)&=\int_0^x\frac{1-\cos(t)}{t}\d t\end{align}$$
I am told that:
$$\ci(x)+\cn(x)=\gamma+\ln(x)$$
I assume dominated convergence for the sake of sum/integral exchange:
$$\begin{align}\ci(x)+\cn(x)&=\lim_{N\to\infty}\left[\int_0^x\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n)!}\cdot t^{2n-1}\d t+\int_x^N\frac{-1}{t}+\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n)!}\cdot t^{2n-1}\d t\right]\\&=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n)!}\cdot\frac{x^{2n}}{2n}+\ln(x)\lim_{N\to\infty}\left[-\ln(N)+\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n)!}\cdot\frac{N^{2n}-x^{2n}}{2n}\right]\\&=\ln(x)+\lim_{N\to\infty}\left[-\ln(N)+\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n)!}\cdot\frac{N^{2n}}{2n}\right]\end{align}$$
Empirically the limit on the right does in fact tend to $\gamma$, but I can see no way to resolve it; applying Maclaurin's series on $\ln(1/N)$ doesn't help, attempting to recast the sum as an integral doesn't help, and for this to converge to $\gamma$ in fact the sum must converge to the harmonic sum – but how? I am roundly stuck. I also expect that the canonical way of resolving the $\ci(x)+\cn(x)$ identity is more elegant than this, but I have not been able to find any resources on the matter.
How should I proceed?
Note: My question really boils down to showing that:
$$\lim_{N\to\infty}\sum_{n=1}^\infty\left[\frac{1}{n}-\frac{(-1)^{n+1}}{(2n)!}\cdot\frac{N^{2n}}{2n}\right]=0$$
Which Desmos beautifully corroborates (until the numbers grow too large to handle) so I am confident this limit is correct: I just have no idea how to show it!
Best Answer
Since trigonometric functions are essentially complex exponentials, one is motivated to directly study the behavior of
\begin{aligned} I(x) &=\int_0^x{1-e^{it}\over t}\mathrm dt-\int_x^\infty{e^{it}\over t}\mathrm dt \\ &=\int_0^{-ix}{1-e^{-u}\over u}\mathrm du-\int_{-ix}^{-i\infty}{e^{-u}\over u}\mathrm du \end{aligned}
where $x$ is a positive real number. To handle the first integral, we can use Cauchy's theorem to conclude
$$ \int_0^{-ix}{1-e^{-u}\over u}\mathrm du=\int_0^1{1-e^{-u}\over u}\mathrm du+\log(-ix)-\int_1^{-ix}{e^{-u}\over u}\mathrm du $$
Plugging this back in, we have
$$ I(x)=\log(-ix)+\int_0^1{1-e^{-u}\over u}\mathrm du-\left(\int_1^{-ix}+\int_{-ix}^{-i\infty}\right){e^{-u}\over u}\mathrm du $$
Now it remains to simplify the latter contour integral. Again, we can use Cauchy's theorem to conclude that it is equal to
$$ \int_1^{-i\infty}{e^{-u}\over u}\mathrm du=\lim_{T\to+\infty}\int_1^{-iT}{e^{-u}\over u}\mathrm du $$
To compute the complex integral, we construct a fan-shape contour so that the segment $[1,-iT]$ is joined by the segument $[1,T]$ and an arc $C_T$ connecting $-iT$ and $T$. Applying Cauchy's theorem, we get
$$ \int_1^{-iT}{e^{-u}\over u}\mathrm du=\left(\int_1^T{e^{-u}\over u}+\int_{C_T}\right){e^{-u}\over u}\mathrm du $$
If we write $u=Te^{i\theta}$, then we can see that on $C_T$ the integrand satisfies
$$ \left|{e^{-u}\over u}\right|\le{e^{-T\cos\theta}\over T} $$
Now, by mean value theorem for integrals we conclude that there exists $\varepsilon>0$ such that
$$ \int_{C_T}{e^{-u}\over u}\mathrm du=\mathcal O(e^{-\varepsilon T}) $$
Letting $T\to+\infty$, we obtain
$$ I(x)=\log(-ix)+\underbrace{\int_0^1{1-e^{-u}\over u}\mathrm du-\int_1^\infty{e^{-u}\over u}\mathrm du}_\gamma $$
Taking real components on both sides gives the desire conclusion. Hope this proof can help!