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How to find a positive function $g(x)$ such that \begin{align} \int_0^\infty e^{-\frac{c}{2x^2}} g(x) dx =\sqrt{\frac{\pi}{2}} e^{-\frac{\sqrt{c}}{2}} \end{align}

I know that the solution is $e^{-\frac{x^2}{2}}$ but not sure how to search for such a solution.

Boby
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  • Integrate by parts maybe? – user217285 Dec 29 '16 at 19:37
  • @Nitin Not sure if this will work. – Boby Dec 29 '16 at 19:45
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    There are many such functions. One is the constant function $$g(x)=\frac{\sqrt{\pi/2}e^{-\sqrt{c}/2}}{\int_0^{\infty}e^{-c/2t^2}dt}$$ – MPW Dec 29 '16 at 19:45
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    @MPW You are clearly correct by saying that there are infinitely many such functions, but the one that you wrote is not one of them since it is undefined: note that the integral in the denominator does not converge. – yohBS Dec 29 '16 at 19:52
  • Let $x = \frac{1}{t}$, and pick $g$ to cancel an unpleasant factor. – user58697 Dec 29 '16 at 20:26
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    Hint: by changing the variable $x=1/\sqrt(t}$, your integral becomes the Laplace transform of a new function. It can be inverted using standard tables and you obtain the result. – Paul Enta Dec 29 '16 at 20:44
  • @yohBS : Ah. Well, I was trying to be a smart-aleck and put my foot in my mouth. :( Touche, +1. – MPW Dec 29 '16 at 21:03

1 Answers1

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A little guesswork leads to the ansatz

$$g=b \exp \left(-a x^2\right)$$

Inserting into the integral gives for the two parameters

$$a=\frac{1}{8}, b=\frac{1}{2}$$

This is a soluton. I don't know if there are more. The comment of Paul Enta relating the Problem to Laplace Transformation suggests that there is only one positive solution.

If we drop the condition of positivity for $g$ we can find more solutions.

I tried to look for a function g which makes the integral equal to zero. A good candidate could be something convergent with a sine function.

The ansatz

$$\int_0^{\infty } \sin \left(p x^2\right) \exp \left(-\frac{c}{2 x^2}\right) \, dx$$

is successful with

$$p=\frac{\left(\frac{3 \pi }{4}\right)^2}{c}$$

Another zero solution comes with the cosine

$$\int_0^{\infty } \exp \left(-\frac{c}{2 x^2}\right) \cos \left(\frac{\left(\frac{\pi x}{4}\right)^2}{c}\right) \, dx=0$$

Dr. Wolfgang Hintze
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  • You're right, there are other solutions by adding contributions as the ones you show. The Laplace method I proposed gives the only solution $g(t)$ such that $t^{-3/2}g(t^{-1/2})$ can be Laplace transformed. – Paul Enta Dec 29 '16 at 21:30
  • Yes, and, as I said, for positive g there is hence only this one. – Dr. Wolfgang Hintze Dec 29 '16 at 21:34
  • Not sure if the Ansatz : $g(x)=e^{-ax^2}f(x)$ can be made f any integrable function. Then the integral gives a number, thus adding a constant to f can be done to reach the desired value ? – QuantumPotatoïd May 03 '21 at 10:31