Thread: series expansion of the cosine integral

1. series expansion of the cosine integral

$\displaystyle \text{Ci} (x) = -\int_{x}^{\infty} \frac{\cos t}{t} \ dt$

$\displaystyle = -\int_{x}^{\infty} \frac{1}{t} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n!)} \ t^{2n} \ dt$

$\displaystyle = -\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{x}^{\infty} t^{2n-1} \ dt$ (since the series expansion of the cosine function converges uniformly)

$\displaystyle = -\int^{\infty}_{x} \frac{dt}{t} -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{x}^{\infty} t^{2n-1} \ dt$

$\displaystyle = -\ln t\Big|^{\infty}_{x} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ t^{2n} \Big|_{x}^{\infty}$

$\displaystyle \lim_{t \to \infty} \Big(-\ln t - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} t^{2n} \Big) + \ln x + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ x^{2n}$

I thought this had to be incorrect, but Maple says that the limit does indeed evaluate to Euler's Constant. Why?

2. Re: series expansion of the cosine integral

Originally Posted by Random Variable
$\displaystyle \text{Ci} (x) = -\int_{x}^{\infty} \frac{\cos t}{t} \ dt =$

...

$\displaystyle = \lim_{t \to \infty} \Big(-\ln t - \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} t^{2n} \Big) + \ln x + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (2n)!} \ x^{2n}$

I thought this had to be incorrect, but Maple says that the limit does indeed evaluate to Euler's Constant. Why?
The problem is the computation of the limit...

$\displaystyle \lambda=\lim_{t \rightarrow \infty} \{\ln t + \int_{0}^{t} \frac{\cos \tau-1}{\tau}\ d\tau\}$ (1)

Applying a fundamental property of the Laplace Transform first You compute [I obtained the esult some time ago]...

$\displaystyle \mathcal{L}\{\frac{\cos \tau-1}{\tau} \}= \int_{s}^{\infty}\frac{1}{\sigma\ (1+\sigma^{2})}\ d \sigma = \frac{1}{2}\ \ln \frac{1+s^{2}}{s^{2}}$ (2)

Then You find...

$\displaystyle \mathcal{L}\{\int_{0}^{t} \frac{\cos \tau-1}{\tau} \}= \frac{1}{2 s}\ \ln \frac{1+s^{2}}{s^{2}} = \frac{1}{s}\ \{\ln \sqrt{1+s^{2}} - \ln s\}$ (3)

Third step is the application of the 'final value theorem' [remember that is $\displaystyle \mathcal{L}\{\ln t \}= - \frac{\gamma- \ln s}{s}$ obtaining...

$\displaystyle \lambda= \lim_{s \rightarrow 0} \{- \gamma + \ln s + \ln \sqrt{1+s^{2}} - \ln s \}= - \gamma$ (4)

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$

3. Re: series expansion of the cosine integral

What do we need to show to justify the use of the FVT?

EDIT: Do we just need to show that the the original limit exists?

4. Re: series expansion of the cosine integral

Originally Posted by Random Variable
What do we need to show to justify the use of the FVT?

EDIT: Do we just need to show that the the original limit exists?
The final value theorem exptablishes that $\displaystyle \lim_{t \rightarrow \infty} h(t)= \lim_{s \rightarrow 0} s\ H(s)$ can be applied if $\displaystyle H(s)$ has no poles with real part greater than 0 and no poles in $\displaystyle s=i\ \omega$ with $\displaystyle \omega \ne 0$...

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$