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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!

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1 Answer 1

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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?

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  • $\begingroup$ The only term that is not matching is for $\lambda$. @mark $\endgroup$
    – Atul Anurag Sharma
    Commented Oct 18, 2020 at 14:12
  • $\begingroup$ $$F(\lambda) = \int_{0}^{\infty} \frac{1}{\lambda}e^{- \lambda t}\ cos(t) \ dt \sim \frac{1}{\lambda} \sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} \frac{\Gamma(n+1)}{\lambda^{n+1}} \\ \sim \frac{1}{\lambda} \sum_{n=0,2,4, \cdot \cdot}^{\infty} \frac{(-1)^{n/2}}{\lambda^{n+1}} \\ \sim \frac{1}{\lambda} \sum_{n=0 }^{\infty} \frac{(-1)^{n}}{\lambda^{2n+1}} \\ \sim \frac{1}{\lambda^2} \sum_{n=0 }^{\infty} \frac{(-1)^{n}}{\lambda^{2n}} $$ $\endgroup$
    – Atul Anurag Sharma
    Commented Oct 18, 2020 at 14:32
  • $\begingroup$ All terms match now. For the first series you wrote replace $\lambda^{n+1}$ with $\lambda^{2n+1}$. $\endgroup$
    – Mark Viola
    Commented Oct 18, 2020 at 14:34
  • $\begingroup$ Yes. You have it now! Well done. $\endgroup$
    – Mark Viola
    Commented Oct 18, 2020 at 14:36
  • $\begingroup$ Apart from that, everything in my process is right? $\endgroup$
    – Atul Anurag Sharma
    Commented Oct 18, 2020 at 14:37

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