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$A, B, C > 0$, $x$ is complex and $Re(x)>0$. My guess is that $f(x)=0$ but I don't know how to prove it.

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    $\begingroup$ do you have any hypothesis about $f$, continuous, differentiable, analytic...? $\endgroup$ – Marsan Feb 8 at 20:26
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    $\begingroup$ Unfortunately, no. $f(x)$ is actually the Laplace transform of a function I am trying to find. I originally had an integral equation which included convolutions so I took the Laplace transform of the whole thing in the hopes of solving it. The resulting equation had some terms in the RHS as well. I was hoping to use an approach similar to that used for differential equations by first solving the resulting equation with RHS equal to 0, which is what I posted here. $\endgroup$ – Zaeem Hussain Feb 11 at 16:22
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For $f(x)\left(A-\dfrac{B}{x+C}\right)+Df(x+C)=0$ ,

$Df(x+C)=\dfrac{B-A(x+C)}{x+C}f(x)$

$f(Cx+C)=\dfrac{B-A(Cx+C)}{D(Cx+C)}f(Cx)$

$f(C(x+1))=\dfrac{B-AC(x+1)}{CD(x+1)}f(Cx)$

$f(C(x+1))=\dfrac{-AC\left(x-\dfrac{B}{AC}+1\right)}{CD(x+1)}f(Cx)$

$f(C(x+1))=-\dfrac{A}{D}\dfrac{x-\dfrac{B}{AC}+1}{x+1}f(Cx)$

With reference to http://eqworld.ipmnet.ru/en/solutions/fe/fe1105.pdf,

The general solution is $f(Cx)=\Theta_1(x)\dfrac{A^x\Gamma\left(x-\dfrac{B}{AC}+1\right)}{D^x\Gamma(x+1)}$, where $\Theta_1(x)$ is an arbitrary unit antiperiodic function

$f(x)=\Theta(x)\dfrac{A^\frac{x}{C}\Gamma\left(\dfrac{x}{C}-\dfrac{B}{AC}+1\right)}{D^\frac{x}{C}\Gamma\left(\dfrac{x}{C}+1\right)}$, where $\Theta(x)$ is an arbitrary antiperiodic function with period $C$

Similarly, for $f(x)\left(A-\dfrac{B}{x+\dfrac{B}{A}}\right)+Cf\left(x+\dfrac{B}{A}\right)=0$ ,

The general solution is $f(x)=\Theta(x)\dfrac{A^\frac{Ax}{B}\Gamma\left(\dfrac{Ax}{B}\right)}{C^\frac{Ax}{B}\Gamma\left(\dfrac{Ax}{B}+1\right)}=\Theta(x)\dfrac{A^{\frac{Ax}{B}-1}B}{C^\frac{Ax}{B}x}$, where $\Theta(x)$ is an arbitrary antiperiodic function with period $\dfrac{B}{A}$

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  • $\begingroup$ Thanks, but I think you cannot apply the result in eqworld.ipmnet.ru/en/solutions/fe/fe1105.pdf directly for $f(Cx)$. If we change the variables to $y=Cx$, we get an equation of the form $f(y+C)=R(y)f(y)$, whereas the result in the link only applies if $f(y+1)=R(y)f(y)$. Do you know of a more general result that would apply to $f(y+C)=R(y)f(y)$? $\endgroup$ – Zaeem Hussain Apr 17 at 15:56

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