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I'm doing some fouriertransform, and can't get the last step correct when Im comparing to an old exam with the question.

Its the fourier transform of $f(x)=\frac{1}{1+7x^2}$.

Where my solution is wrong is when I compute the actual residue.

My solution, a couple of steps in:

$f(z)=\frac{1}{1+7z^2} \Rightarrow z= \pm \frac{i}{\sqrt{7}}$ where $\frac{i}{\sqrt{7}} \in int(\gamma_R)$ and the residue we want to calculate.

I would like to say that the residue is $\lim_{z \rightarrow \frac{i}{\sqrt{7}}} (z-\frac{i}{\sqrt{7}}) \frac{e^{-izt}}{(z-\frac{I}{\sqrt{7}}(z+\frac{I}{\sqrt{7}})}$

This is the step that I cant get right, and its just simple algebra...

My answer of above is $\frac{\sqrt{7}e^{\frac{t}{\sqrt{7}}}}{2i}$

While the correct answer according to the exam $\frac{e^{\frac{t}{\sqrt{7}}}}{2i\sqrt{7}}$

Can someone show me the actual steps on how this is correct, I know its a rather simple question...

uoiu
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2 Answers2

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Here's how to find the residue at $i/\sqrt 7$ $$\frac{1}{7z^2+1}=\frac{1}{7(z-i\sqrt 7)(z+i\sqrt 7)}$$ $$=\frac{1}{7(z-i/\sqrt 7)((z-i/\sqrt 7)+2i/\sqrt 7)}$$ $$=\frac{1}{7(z-i/\sqrt 7)(2i/\sqrt 7)(1+(\sqrt 7/2i)(z-i/\sqrt 7))}$$ $$=\frac{-\sqrt 7\space i}{14(z-i/\sqrt 7)(1-(\sqrt 7\space i/2)(z-i/\sqrt 7))}$$ $$=\frac{-\sqrt7 \space i(1+(\sqrt 7\space i/2)(z-i/\sqrt 7)+...)}{14(z-i/\sqrt 7)}$$

P. Lawrence
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Let $$f(z)=\frac{1}{1+7z^2}$$

Then \begin{eqnarray} \frac{1}{1+7z^2} &= \frac{1}{7}\frac{1}{(\frac{i}{7}+z^2)}\\ &= \ldots \end{eqnarray}

When you then apply your residue you have \begin{eqnarray} \text{Res}_{z=i/\sqrt{7}} & =\frac{1}{7}\frac{e^{t/\sqrt{7}}}{2i/\sqrt{7}}\\ &= \frac{\sqrt{7}}{14i}e^{t/\sqrt{7}}\\ &=\frac{1}{2i\sqrt{7}}e^{t/\sqrt{7}}\\ \end{eqnarray}

asymptotic
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  • Thanks! Looking at your solution I realise that, when finding roots, I simply divided by 7 when $7z^2=-1$ without multiplying it back, which gave me the right roots but the 7 disappeared. Thanks! – uoiu Nov 18 '22 at 14:03
  • @zzz__ Precisely, easy mistake! – asymptotic Nov 18 '22 at 14:32