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!



Is this inverted integral transform valid?
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


$$G(iq)=\int\_{0}^{\infty}g(x)J\_{0}(xq)xdx$$


This is a [Hankel transform.](http://en.wikipedia.org/wiki/Hankel_transform) The inverse is


$$g(x)=\int\_{0}^{\infty}G(iq)J\_{0}(xq)qdq$$


