MHB Fourier Transform limits problem

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The discussion focuses on finding the Fourier Transform of the function e^{-a|t|}Cos(bt). The initial approach involves simplifying the cosine term using Euler's formula and splitting the integral based on the absolute value of t. Participants explore the implications of taking the real part of the integral and whether this aligns with the Fourier Cosine Transform. There is a consideration of how to handle the limits of integration and the behavior of the exponential terms. The conversation concludes with a request for clarification on the integration process and its relation to the Fourier Sine Transform.
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Find the Fourier Transform of $ e^{-a|t|}Cosbt $

I'd like to simplify this using $Cosbt = Re\left\{e^{ibt}\right\}$

$\therefore \hat{f}(\omega) = Re\left\{ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\left(-a+ib+iw\right)|t|} \,dt \right\} = Re\left\{ \frac{1}{\sqrt{2\pi}} \frac{1}{-a+ib+iw} e^{-a|t|}.e^{i(b+\omega)|t|}|^\infty_{-\infty} \right\}$

I think I can argue that the $e^{-a|t|}$ term dominates the $e^{i}$ term which is bounded. BUT the lower limit will make
$e^{-a|t|}$ infinite... is there a better way of approaching this FT?
 
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You should recall that
$$|t|=\begin{cases}\phantom{-}t,\; t\ge 0 \\ -t,\; t<0\end{cases}.$$
Then split up your integral into two pieces depending on where your integrand changes.
 
Ackbach said:
You should recall that
$$|t|=\begin{cases}\phantom{-}t,\; t\ge 0 \\ -t,\; t<0\end{cases}.$$
Thanks Ackbach and yes I should!

I'd appreciate if you'd please check the rest ... and sorry about the edits if we cross over.

What I did below seems wrong on reflection, I think I should take the real part only w.r.t. b - if that is possible?

If I did that, would the real part w.r.t. $\omega$ be the same as the Fourier Cosine Transform? And would the Fourier Sine Transform be just Imaginary part of that exponential Transform?

$$\hat{f}(\omega)= \frac{1}{\sqrt{2\pi}} Re\left\{ \int_{-\infty}^{0} e^{at}e^{eibt}e^{i\omega t} \,dt +
\int_{0}^{\infty} e^{at}e^{eibt}e^{i\omega t} \,dt \right\} $$

$$ =\frac{1}{\sqrt{2\pi}} Re\left\{ \frac{1}{a+i(b+\omega)} e^{t(a+i(b+\omega))} |^0_{-\infty} + \frac{1}{-a+i(b+\omega)} e^{t(-a+i(b+\omega))} |^\infty_0 \right\} $$

$$ = \frac{1}{\sqrt{2\pi}} Re\left\{ \frac{1}{a+i(b+\omega)} - \frac{1}{-a+i(b+\omega)} \right\}$$

$$= \frac{1}{\sqrt{2\pi}} \frac{2a}{a^2+(b+\omega)^2} ?$$

Also - would this be the same as the Fourier Cosine Transform? And would the Fourier Sine Transform be just Imaginary part of the exponential Transform?
 
Last edited:
Hi guys, what I would like to know is - can I do the FT using Euler's form for Cos(bt) - and then extract the real part w.r.t. b but leaving the complex parts w.r.t. wt? Or is the only possible approach here to use integration by parts?
 
I think your basic strategy is fine, but there are some details that may need improving:

\begin{align*}\hat{f}(\omega) &= \text{Re}\left\{ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\left(ib+iw\right)t} e^{-a|t|} \,dt \right\} \\
&=\frac{1}{\sqrt{2\pi}}\:\text{Re}\left\{ \int_{-\infty}^{0}e^{\left(ib+iw\right)t} e^{at} \,dt+
\int_{0}^{\infty}e^{\left(ib+iw\right)t} e^{-at} \,dt \right\} \\
&=\frac{1}{\sqrt{2\pi}}\:\text{Re}\left\{ \int_{-\infty}^{0}e^{\left(a+ib+iw\right)t} \,dt+
\int_{0}^{\infty}e^{\left(-a+ib+iw\right)t}\,dt \right\}.\end{align*}

Can you finish from here?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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