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We are given that: $\mathscr{F}_s\{f(x)\}=F_s(k)=\frac k{k^2+1}.$ We need to find: $f(x)=\mathscr{F}_s^{-1}\{F_s(k)\}.$

My work so far:

We have: $$f(x)=\sqrt{\frac2π}\int_0^{\infty}\frac k{k^2+1}\cdot\sin kx\,dk.$$ Therefore \begin{align} f'(x)&=\sqrt{\frac2π}\int_0^{\infty}\frac{k^2}{k^2+1}\cdot\cos kx\, dk\\\\ &=\sqrt{\frac2π}\int_0^{\infty}\cos kx\,dk-\sqrt{\frac2π}\int_0^{\infty}\frac{\cos kx}{k^2+1}\,dk\\\\ &=-\sqrt{\frac2π}\int_0^{\infty}\frac{\cos kx}{k^2+1}\,dk. \end{align} [Since $\int_0^{\infty}\cos t\,dt=0.$ Therefore $\int_0^{\infty}\cos kx\,dk=\frac1x\int_0^{\infty}\cos t\,dt=0$ via the substitution: $kx=t.$]

So $$f''(x)=\sqrt{\frac2π}\int_0^{\infty}\frac{k}{k^2+1}\cdot\sin kx\,dk=f(x).$$We thus have the following ODE:

$(D^2-1)f(x)=0,$ where $D\equiv\frac d{dx}.$ This gives

$f(x)=Ae^x+Be^{-x}.$

Now we have: $f(0)=0\Rightarrow A+B=0$

and $$f''(0)=-\sqrt{\frac2π}\int_0^{\infty}\frac{dk}{k^2+1}=-\sqrt{\fracπ2}.$$ This implies $A-B=-\sqrt{\fracπ2}.$ On solving, we obtain: $A=-\frac12\sqrt{\fracπ2}$ and $B=\frac12\sqrt{\fracπ2}.$ Hence

$\color{green}{f(x)=\frac12\sqrt{\fracπ2}(e^{-x}-e^x).}\tag*{}$

But the answer in my book is given as $$\color{red}{f(x)=\sqrt{\fracπ2}e^{-x}.}$$

Then I also derived this result. For this, I split $f(x)$ as follows: \begin{align} f(x)&=\sqrt{\frac2π}\int_0^{\infty}\frac{k^2}{k(k^2+1)}\cdot\sin kx\,dk\\\\ &=\sqrt{\frac2π}\int_0^{\infty}\frac{\sin kx}{k}\,dk-\sqrt{\frac2π}\int_0^{\infty}\frac{\sin kx}{k(k^2+1)}\,dk\\\\ &=\sqrt{\fracπ2}-\sqrt{\frac2π}\int_0^{\infty}\frac{\sin kx}{k(k^2+1)}\,dk. \end{align} [Since $\int_0^{\infty}\frac{\sin t}t\,dt=\fracπ2.$ Therefore $\int_0^{\infty}\frac{\sin kx}k\,dk=\fracπ2$, via $t=kx.$]

So that we obtain: $f(0)=\sqrt{\fracπ2}$ and $f'(0)=-\sqrt{\fracπ2}$ implying that $A=0$ and $B=\sqrt{\fracπ2}.$ And thus we obtain the above mentioned result (in red).

So my question is, what is wrong with my first approach?

Dhrubajyoti Bhattacharjee
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    The formula $\int_{0}^{+\infty}\cos\left(t\right)dt=0$ is clearly false. – Marco Cantarini Aug 26 '20 at 10:55
  • I agree that the improper integral is not Riemann integrable (or diverges). But please refer to this: https://math.stackexchange.com/q/3803142/762849 – Dhrubajyoti Bhattacharjee Aug 26 '20 at 11:16
  • I think since for the first result (in green): as $x\to\infty,, f(x)\to-\infty$ and as $x\to-\infty,, f(x)\to\infty.$ So $f(x)$ is not a good function for Fourier transform. Am not exactly sure tho.. – Dhrubajyoti Bhattacharjee Aug 26 '20 at 11:23
  • In the linked answer it is shown that the Laplace transform of $\cos (x)$ at $0$ is $0$, but the passage of the limit into the integral cannot be justified, unless one passes to more general objects such as, for example, distributions (and I am not sure it makes sense also in this situation). Certainly the integral, in the classical sense, is not $0$. – Marco Cantarini Aug 26 '20 at 13:45
  • Ok sir.. Got it. Thank you. – Dhrubajyoti Bhattacharjee Aug 26 '20 at 15:57

0 Answers0