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We're given the ODE $$2(x^2+1)y'(x) + (x-i)y(x)=0$$

Deriving $f$ in respect to $x$, we get the equality $$f'(x)=\frac{-1}{2(x+i)}f(x)$$ which lines up nicely as a solution to the above ODE.

All the solutions to this ODE hold the form $$y(x)=\lambda e^{A(t)}$$

with $\lambda\in\mathbb{R}$ fixed and $A(t)$ a fixed anti-derivative of, in this case, $x\mapsto\frac{-1}{2(x+i)}$, so we choose $A(t)=\frac{-1}{2}\ln|x+i|$.

This leaves us with

$$f(x)=\lambda e^{\frac{-1}{2}\ln|x+i|}$$

The problem is that it's nowhere near the expected answer of

$$f(x)=\sqrt{\pi} (x^2+1)^\frac{-1}{4}e^{\frac{i}{2}\arctan{x}}$$

I can also try to directly compute $f(x)$ since the paper suggests to use the Gaussian integral. A quick variable change of $t=u^2$ leaves us with

$$f(x)=2 \int_0^{+\infty} e^{-u^2} e^{i u^2 x} du$$

But I can't get through this integral in order to get $\int_0^{+\infty} e^{-u^2} du$ to appear, can you provide some hints?

Exercise To The Reader
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    I don't follow all of this, but two comments which may help. In the integral for f(x) near the bottom, x is a constant, so you have $\int e^{-ku^2} du$ for some k, which is what I think you want. Secondly, this relationship may help: https://math.stackexchange.com/questions/414248/why-does-arctanx-frac12i-log-left-fracx-ixi-rightk – Blitzer Dec 31 '22 at 15:56
  • @Blitzer Yes, I have $\int_0^{+} e^{(xi-1)u^2} du$, but evaluating it directly leaves me with $$\left[\frac{1}{2u(xi-1)}e^{(xi-1)u^2\right]_0^{+\infty}$$ which cannot be computed at $u=0$, and its limit from the right apparoaches $+\infty$. – Exercise To The Reader Dec 31 '22 at 16:09
  • Are you sure the $i$ is the imaginary unit? Wolfram gives the required answer if $i$ is taken as a parameter – Vincenzo Tibullo Dec 31 '22 at 17:19
  • @enzotib Yes $i$ is the imaginary unit. Later in the problem, I need to show that the required answer is also the Fourier transform of $t\mapsto\frac{e^{-t}}{\sqrt{t}}\mathbb{1}{\mathbb{R}+}(t)$. – Exercise To The Reader Dec 31 '22 at 17:31
  • @ExerciseToTheReader Your evaluation in the comment above is incorrect - $\int_0^{\infty} \exp(-k u^2) du $ is $\sqrt{\pi}/(2 \sqrt{ k}) $. The integrand does not have a simple primitive. I think you used the result for $\exp(-k u)$. – Tom Dickens Dec 31 '22 at 17:47

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A lot has been said in the post as well as in the comments, that is why I'll sum up for the sake of readability. One would like to compute the following Fourier transform : $$ \mathfrak{F}\left[\frac{e^{-t}}{\sqrt{t}}H(t)\right](x) = \int_0^\infty \frac{e^{-t}}{\sqrt{t}}e^{itx}\,\mathrm{d}t =: f(x) $$ where $H(t)$ is the Heaviside function.

The direct way is the substitution $t=u^2$, giving $$ f(x) = 2\int_0^\infty e^{-(1-ix)u^2}\,\mathrm{d}u = \int_{-\infty}^\infty e^{-(1-ix)u^2}\,\mathrm{d}u = \sqrt{\frac{\pi}{1-ix}} $$ where we used firstly the fact that the integrand is even to extend the domain of integration and secondly the gaussian integral formula $-$ which is still valid for complex coefficients !

Another way is differentiation followed by integration by parts : $$ f'(x) = \int_0^\infty it\frac{e^{-t}}{\sqrt{t}}e^{itx}\,\mathrm{d}t = i\int_0^\infty \sqrt{t}\,e^{-(1-ix)t}\,\mathrm{d}t = \left.-\frac{i\sqrt{t}}{1-ix}e^{-(1-ix)t}\right|_{t=0}^{t=\infty} + i\int_0^\infty\frac{1}{2\sqrt{t}}\frac{e^{-(1-ix)t}}{1-ix}\mathrm{d}t = -\frac{f(x)}{2(x+i)}, $$ which gives a first-order linear ODE, whose solution is given by $$ f(x) = A\exp\left(-\frac{1}{2}\ln(x+i)\right) = \frac{A}{\sqrt{x+i}} $$ with the initial condition $$ f(0) = \int_0^\infty \frac{e^{-t}}{\sqrt{t}}\,\mathrm{d}t = 2\int_0^\infty e^{-u^2}\,\mathrm{d}u = 2\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u = \sqrt{\pi} = \frac{A}{\sqrt{i}} $$ hence $A=\sqrt{i\pi}$ and $ f(x) = \sqrt{\frac{\pi}{1-ix}}$ as with the direct calculation.

Finally, knowing that $\arctan(x) = -i\mathrm{arctanh}(ix) = -\frac{i}{2}\ln\left(\frac{1+ix}{1-ix}\right)$, one checks easily that $$ \frac{\sqrt{\pi}}{(1+x^2)^{1/4}}\exp\left(\frac{i}{2}\arctan(x)\right) = \frac{\sqrt{\pi}}{(1+x^2)^{1/4}}\left(\frac{1+ix}{1-ix}\right)^{1/4} = \sqrt{\frac{\pi}{1-ix}} = f(x) $$

Abezhiko
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