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I have the following transform:

$$F(y) = \int_{0}^{\infty} y\exp{\left[-\frac{1}{2}(y^2 + x^2)\right]} I_0\left(xy\right)f(x)\;\mathrm{d}x$$

with the following conditions:

  • $f(x)$ and $F(y)$ must be real and positive (they are continuous probability distributions)
  • $x,y$ must be real and positive (they are magnitudes)

@Carlo suggested the following approach. Re-arrange such that we have: $$\frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} = \int_{0}^{\infty} \left(\frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]}\right) I_0\left(xy\right)x\;\mathrm{d}x$$

Now define:

$$ G(y) = \frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} \quad\quad g(x) = \frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]}$$

We can now write: $$G(y) = \int_{0}^{\infty} g(x) I_0\left(xy\right)x\;\mathrm{d}x$$

Make the substitution $y = iq$ along with the fact that $I_0(ixq) = J_0(-xq) = J_0(xq)$ and we have:

$$G(iq) = \int_{0}^{\infty} g(x) J_0\left(xq\right)x\;\mathrm{d}x$$

This is a Hankel transform, which has an inverse: $$g(x)=\int_{0}^{\infty}G(iq)J_{0}(xq)q\;\mathrm{d}q$$

Substituting in $q = \frac{1}{i}y$ and $\mathrm{d}q = \frac{1}{i}\mathrm{d}y$ we obtain:

$$g(x)= -\int_{0}^{i\infty}G(y)I_{0}(xy)y\;\mathrm{d}y$$

Substituting out $G(y), \;g(x)$:

$$\frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]} = -\int_{0}^{i\infty} \frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} I_{0}(xy)y\;\mathrm{d}y$$

and finally solve for $f(x)$:

$$f(x) = -\int_{0}^{i\infty} x\exp{\left[\frac{1}{2}(y^2 + x^2)\right]} I_0\left(xy\right)F(y)\;\mathrm{d}y$$

  • Is this inversion valid given the aforementioned conditions?

  • How can I perform this integral numerically?

I'm directly measuring $F(y)$ (by normalising the histogram of a large set of values) and so I need to calculate $f(x)$ by doing this integral numerically. However I don't understand how to do this when the data is all real but the integral is over the imaginary axis. Can anyone help?

Thank you!

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up vote 6 down vote accepted

define $G(y)=y^{-1}\exp(y^2/2)(Tf)(y)$ and $g(x)=x^{-1}\exp(-x^2/2)f(x)$, then you seek the solution to the integral equation


This is a Hankel transform. The inverse is


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Thanks for your answer! although in the two integrals you have $f(x)$ when i think you mean $g(x)$? I've edited my question to include the derivation you suggest, along with the extra steps needed to get $f(x)$. Would you mind checking my logic is sound? Thanks again. – CBowman Oct 11 '13 at 1:45
thanks for spotting the typo, I corrected it; with respect to your update: when you change variables in the integral from $q$ to $y=iq$, the integral over $y$ would then run along the positive imaginary axis. – Carlo Beenakker Oct 11 '13 at 7:27
Ah I see, I've never studied integration in the complex plane; what is the appropriate notation to denote that it is along the positive imaginairy axis? thanks – CBowman Oct 11 '13 at 9:43
the way to think about this, is that the Hankel transform with $J_0$ is like a Fourier transform, whereas your transform with $I_0$ is like a Laplace transform. The inversion formula for the Laplace transform also involves an integration along the imaginary axis $\int_{0}^{i\infty}dy$ – Carlo Beenakker Oct 11 '13 at 9:46
That's a help picture to have in mind, thanks again! I'll change the upper limit to $i\infty$ where needed. – CBowman Oct 11 '13 at 9:51

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