Watson's Lemma and Laplace transformation

Question

Use Watson's lemma to find the expansion of $$F(\lambda) = \int_{0}^{\infty} e^{- \lambda t} sin(t)dt$$ ,and verfify the answer using Laplace transformation by expanding the answer using Taylor's expansion.

Comparing the above integral with $$F(\lambda) = \int_{0}^{\infty} e^{- \lambda t} f(t)dt.$$

For applying Watson's lemma we need $f(0) \neq 0$, in our case it is $f(0)=0.$ So I integrated by parts before applying Watson's Lemma: $$F(\lambda) = \int_{0}^{\infty} \frac{1}{\lambda}e^{- \lambda t}\ cos(t) \ dt \sim \frac{1}{\lambda} \sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} \frac{\Gamma(n+1)}{\lambda^{n+1}} \sim \frac{1}{\lambda} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\lambda^{n+1}} $$

Using Laplace transform: $$F(\lambda) = \int_{0}^{\infty} e^{- \lambda t} sin(t)dt = \frac{1}{1+ \lambda^{2}} = \frac{1}{\lambda^{2}} \sum_{n=0}^{\infty} \frac{(-1)^n}{\lambda^{2n}}$$

My Question

I am not able to figure out where I am making mistakes as I am getting different answers with two different methods. Can someone figure out my mistake?

Thanks for the help!

score 1 · Accepted Answer · answered Oct 18 '20 at 04:28

1

If $f(t)=\cos(t)$, then

$$\begin{align} f'(t)&=-\sin(t)\\\\ f''(t)&=-\cos(t)\\\\ f'''(t)&=\sin(t)\\\\ f''''(t)&=\cos(t) \end{align}$$

Therefore, we find that $f^{(n)}(0)=0$ for each odd integer $n$ and $f^{(n)}(0)=(-1)^{n/2}$ for each even integer $n$.

Can you apply Watson's Lemma now?

answered Oct 18 '20 at 04:28

Mark Viola

179,405

The only term that is not matching is for $\lambda$. @mark – Atul Anurag Sharma Oct 18 '20 at 14:12
$$F(\lambda) = \int_{0}^{\infty} \frac{1}{\lambda}e^{- \lambda t}\ cos(t) \ dt \sim \frac{1}{\lambda} \sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} \frac{\Gamma(n+1)}{\lambda^{n+1}} \ \sim \frac{1}{\lambda} \sum_{n=0,2,4, \cdot \cdot}^{\infty} \frac{(-1)^{n/2}}{\lambda^{n+1}} \ \sim \frac{1}{\lambda} \sum_{n=0 }^{\infty} \frac{(-1)^{n}}{\lambda^{2n+1}} \ \sim \frac{1}{\lambda^2} \sum_{n=0 }^{\infty} \frac{(-1)^{n}}{\lambda^{2n}} $$ – Atul Anurag Sharma Oct 18 '20 at 14:32
All terms match now. For the first series you wrote replace $\lambda^{n+1}$ with $\lambda^{2n+1}$. – Mark Viola Oct 18 '20 at 14:34
Yes. You have it now! Well done. – Mark Viola Oct 18 '20 at 14:36
Apart from that, everything in my process is right? – Atul Anurag Sharma Oct 18 '20 at 14:37
Yes. Your methodology is correct. And you have the correct result. – Mark Viola Oct 18 '20 at 14:39
Let us continue this discussion in chat. – Atul Anurag Sharma Oct 18 '20 at 14:39

Watson's Lemma and Laplace transformation

1 Answers1