Integral = f(a) + const

Question

Fairly simple question, but I can not think of an example. Let $f_a(x)$ be an integrable function, that depends on some parameter $a$.

I'm wondering if there exists a function such that $$\int_{-\infty}^\infty f_a(x) \, {\rm d}x = f(a) + {\rm const}$$ i.e. that the result has a constant term independent of an $a$-dependent term, of course both $\neq 0$. Thanks

EDIT: Of course I should have said, that simple constructions by addition of an $a$-independent term does not count! What I mean is more something like $$ \int_{-\infty}^{\infty} \frac{\sin(ax)}{x} \, {\rm d}x = \pi \, .$$

Also worth mentioning is, that it can not be of the form $$ \int_{-\infty}^{\infty} f(ax) \, {\rm d}x = \frac{1}{a} \int_{-\infty}^{\infty} f(z) \, {\rm d}z \, . $$

Likewise as the $\sin$ example something like $$ \int_{-\infty}^{\infty} f(ax) \, \frac{{\rm d}x}{x} $$ will always be independent of $a$. Atm I'm tempted to say this is not possible, but I can not proof it :-/

Answer 1

Let $f_a(x) = e^{-|ax|} + e^{-|x|}$, then

$$\int_{-\infty}^{+\infty} (e^{-|ax|} + e^{-|x|}) dx = \frac{2}{|a|} + 2, a \not= 0$$

That satisfies your condition if I understand it correctly.

Answer 2

Of course:

$$\int_{-\infty}^\infty e^{ -\frac{x^2}{2a^2}}+e^{ -\frac{x^2}{2}}dx = a\sqrt{2\pi} + \sqrt{2\pi}$$