# Solve $\int_{0}^{x}f(t)f(x-t)dt=e^{2x}-1,$ where $f(x)$ is continuous and nonnegative, for $f(x)$.

I try to solve an integral equation as follow: $$\int_{0}^{x}f(t)f(x-t)dt=e^{2x}-1, \textrm{where }f(x) \textrm{ is continuous and nonnegative.}$$ I want to solve such $f(x)$，but after trying to take the derivative of both sides of the equation, I still have no idea. The only thing I get is $f(0)=\sqrt{2}$. Anyone can help me with this?

• Actually if we only assume $f$ is continuous, then we can't even take the derivative. So I think if we need to assume $f$ is smooth or something like this? – Bob brant Sep 21 '17 at 4:55
• This equation has a convolutional form $f*f(x)=e^{2x}-1$ under the following restriction: $x>0$. Usually, this kind of equation is solved using Laplace transform, but here I don't see how it could give an explicit solution... – Jean Marie Sep 21 '17 at 5:06

With Laplace transform: $$\mathcal{L}\int_{0}^{x}f(t)f(x-t)dt=\mathcal{L}(e^{2x}-1)$$ $$(\mathcal{L}f)^2=\dfrac{1}{s-2}-\dfrac1s$$ $$\mathcal{L}(f)=\dfrac{\sqrt{2}}{\sqrt{s(s-2)}}$$ $$f(x)=\sqrt{2}e^tI_0(x)$$ where $I_0(t)=J_0(it)$ is Bessel function.