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How to find the Fourier transform of the following function:

$$f(x)=\frac{1}{x^2+6x+13}$$

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5 Answers 5

Breaking down to partial fractions yields $$ \begin{align} \frac1{x^2+6x+13} &=\frac1{(x+3)^2+4}\\ &=\frac1{(x+3-2i)(x+3+2i)}\\ &=\frac1{4i}\left(\frac1{x+3-2i}-\frac1{x+3+2i}\right) \end{align} $$ Therefore, for $\xi\ge0$, $$ \begin{align} \int_{-\infty}^\infty\frac1{x^2+6x+13}e^{-i2\pi x\xi}\,\mathrm{d}x &=\frac1{4i}\int_{-\infty}^\infty\left(\frac1{x+3-2i}-\frac1{x+3+2i}\right)e^{-i2\pi x\xi}\,\mathrm{d}x\\ &=\frac1{4i}\int_{\gamma}\left(\frac1{z+3-2i}-\frac1{z+3+2i}\right)e^{-i2\pi z\xi}\,\mathrm{d}z\\ &=\frac{2\pi i}{4i}e^{-i2\pi(-3-2i)\xi}\\ &=\frac\pi2e^{(-4+6i)\pi\xi}\\ &=\frac\pi2e^{(-4|\xi|+6i\xi)\pi}\quad\text{for all }\xi \end{align} $$ where $\gamma$ is the curve passing left to right along the real axis then circles back clockwise around the lower half-plane.

Thus, the Fourer Transform is $$ \frac\pi2e^{-4\pi|\xi|}(\cos(6\pi\xi)+i\sin(6\pi\xi)) $$

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A related problem. The Fourier transform is given by,

$$ \int_{-\infty}^{\infty}\frac{e^{-ixt}}{x^2+6x+13}dx = \int_{-\infty}^{\infty}\frac{e^{-ixt}}{(x+3)^2+4}dx . $$

Making the change of variables $y=x+3$, we have

$$ \int_{-\infty}^{\infty}\frac{e^{-i(y-3)t}}{y^2+4}dy = e^{3it}\int_{-\infty}^{\infty}\frac{e^{-iyt}}{y^2+4}dy .$$

Now to evaluate the last integral, you can use contour integration by considering the complex integral

$$ \oint_{C}\frac{e^{-itz}}{{z^2+4}}dz, $$

where $z=y+iw$ and $C$ is a suitable contour. See example(II) for details.

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This begs for completing the square: $$ x^2 + 6x + 13 = \Big( x^2 + 6x + 9 \Big) + 4 = (x+3)^2 + 4. $$ Then $$ \frac{1}{(x+3)^2 + 4} = \frac 1 4 \cdot \frac{1}{\left(\dfrac{(x+3)^2}{4}\right) + 1} =\frac 1 4 \cdot \frac{1}{\left(\frac{x+3}{2}\right)^2+1} = \frac 1 4 \cdot \frac{1}{w^2 + 1} $$

The Laplace transform involves an integral with respect to $x$. So $$ w = \frac{x+3}{2} $$ $$ dw = \frac{dx}{2} $$ $$ dx = 2\,dw $$

If you can find $\displaystyle \int_0^\infty \frac{1}{1+w^2} e^{-tw}\,dw$, you should be able to get what you're looking for.


If you'd had a number smaller than $9$ where $13$ appeared, then things would be different. For example: $$ x^2 + 6x + 5 = \Big(x^2+6x+0\Big) - 4 = (x+3)^2 - 2^2 = (x+3+2)(x+3-2) $$ $$ =(x+5)(x+1). $$ Then $$ \frac{1}{x^2+6x+5} = \frac{1}{(x+5)(x+1)} = \frac{A}{x+5}+\frac{B}{x+1}, $$ and you'd have to find $A$ and $B$.

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Or use some practical magic. (I'll consider $g(x) = f(x-3) = \frac{1}{x^2+4}$ instead of $f$.) If $\mathcal{F}$ is the unitary angular frequency Fourier transform then

$$ -\mathcal{F}(g)'' + 4 \, \mathcal{F}(g) = \mathcal{F} \left( (x^2+4) g \right) = \mathcal{F}(1) = \sqrt{2 \pi} \, \delta $$

which together with $$ \mathcal{F}(g)(0) = \tfrac{1}{4} \sqrt{2 \pi}$$ correctly suggests that $$\mathcal{F}(g)(x) = \tfrac{1}{4}\sqrt{2 \pi} \, e^{-2|x|}.$$

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If you have Rudin's "Real and Complex Analysis", look at p. 183 in Chapter 9. As he points out, by directly calculating, the inverse Fourier transform of $e^{-\lambda |\xi|}$ for $\lambda > 0$ is seen to be $$\sqrt{2 \over \pi} {\lambda \over \lambda^2 + x^2}$$ This means the Fourier transform of ${\displaystyle {1 \over 4 + x^2}}$ is given by $$\sqrt{\pi} {e^{-2|\xi|} \over 2\sqrt{2}}$$ Since ${\displaystyle {1 \over x^2 + 6x + 13} = {1 \over (x + 3)^2 + 4}}$, the Fourier transform of ${\displaystyle {1 \over x^2 + 6x + 13}}$ is then $$\sqrt{\pi} {e^{-2|\xi|+3\xi} \over 2\sqrt{2}}$$

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