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From,

$$ F(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(t) \cdot e^{-ixt} \, \Bbb dt $$

I obtained,

$$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{|-t|} \cdot e^{-i \cdot x \cdot t} \, \Bbb dt $$

Consequently I found $F(t)$ as,

$$ \sqrt{\frac{2}{\pi}} \cdot \frac{1}{1 + x^2} $$

Is that the Fourier transform of $f(t)=\exp(-|t|)$, or should I continue with the following formula?

From, $$ f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(x) \cdot e^{ixt} \, \Bbb dx $$

Lastly I found like that but I don’t have any idea about what did I find…

$$ \frac{2}{\pi} \int_{0}^{\infty} \frac{2\cos(xt)}{1 + x^2} \, \Bbb dx $$

I have wondered what is the general solution way to find a Fourier transform of any function with this two formulas?

Thank you everyone.

Rócherz
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LexusLS
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1 Answers1

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Split the integral into two parts: $$F(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-|x|}e^{-ixt}\mathrm{d}x=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{(1-it)x}\mathrm{d}x+\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-(1+it)x}\mathrm{d}x=\sqrt{\frac{2}{\pi}}\frac{1}{1+t^2}$$ is no problem.

The inverse Fourier transform $$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(t)e^{ixt}\mathrm{d}t=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{e^{ixt}}{1+t^2}\mathrm{d}t=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\cos xt}{1+t^2}\mathrm{d}t+\frac{i}{\pi}\int_{-\infty}^{\infty}\frac{\sin xt}{1+t^2}\mathrm{d}t$$ One can get the result quickly by complex contour integral.

If you want to know the Fourier transform of a function $f(x)$, you can use $$F(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ixt}\mathrm{d}x$$

If you already known the Fourier transform $F(t)$, you will use $$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(t)e^{ixt}\mathrm{d}t$$ to find the original function.

Cunyi Nan
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