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I've run into having to solve this equation for $f(x)$:

$$1+ax=\int_{-\infty}^xf(x-t)dt$$

Unfortunately, I am not familiar with solving integral equations. Can anyone help? Is is even soluble?

Edit: Fixed a typo in the upper limit in the integral.

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Is there a typo? Since using the substitution $x-t=s$ in the integral we get that $1+ax$ should be constant and $a=0$. –  Davide Giraudo Feb 25 '12 at 19:18
    
I re-checked my math that led to this equation, and indeed found a typo. Will fix shortly. –  M.B.M. Feb 25 '12 at 20:38
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Unfortunately, the same exact substitution Davide points out shows that the RHS is the constant $$\displaystyle \int_0^\infty f(s)ds.$$ So nothing really changes here... –  anon Feb 25 '12 at 20:47
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1 Answer

up vote 2 down vote accepted

Note that rhs of this equation is constant. Indeed, $$ \int\limits_{-\infty}^{x} f(x-t)dt= \{\tau=x-t\}= \int\limits_{0}^{+\infty}f(\tau)d\tau= \int\limits_{0}^{+\infty}f(t)dt $$ Therefore the lhs of this equation must be constant. But this is possible only if $a=0$. For the case when $a=0$, we have $1=\int\limits_{0}^{+\infty}f(t)dt$, otherwise there is no solution.

Finally if $a=0$ the solution of this equation is any integrable function $f$ such that $\int\limits_{0}^{+\infty}f(t)dt=1$. If $a\neq 0$, solution doesn't exist.

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@DilipSarwate, Bullmoose Corrected everything you pointed out –  Norbert Feb 25 '12 at 20:47
    
Ok, thank you! I think I made a deeper mistake in the problem that led me to this question. I'll dig tomorrow. –  M.B.M. Feb 26 '12 at 5:11
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