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?

Re: series expansion of the cosine integral

Quote:

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)

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

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

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?

Re: series expansion of the cosine integral

Quote:

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 $\displaystyle \omega \ne 0$...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

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

Re: series expansion of the cosine integral

Wouldn't those be the conditions if we didn't know $\displaystyle h(t)$$? Although in this case those conditions are obviously satisfied. The article on Wikipedia suggests that we only need to show that $\displaystyle \lim_{t \to \infty} h(t)$ is finite (which is easier said than done for this problem).

Re: series expansion of the cosine integral

Quote:

Originally Posted by

**Random Variable** Wouldn't those be the conditions if we didn't know $\displaystyle h(t)$$? Although in this case those conditions are obviously satisfied. The article on Wikipedia suggests that we only need to show that $\displaystyle \lim_{t \to \infty} h(t)$ is finite (which is easier said than done for this problem).

What is the article of Wikipedia You talk about?... if is is this one...

Laplace transform - Wikipedia, the free encyclopedia

... it is clearly explained that if h(t) has finite limit, then the FVT supplies this limit only if $\displaystyle s\ H(s)$ has all poles in the left half plane. If the article is another, then it is possible that something different is written because in my opinion Wikipedia is not completely reliable...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

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