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Can we find on $\mathbb{R}^+$ a real positive function $f(x)$ (in $C^{\infty}$) such that:

$$\int_0^{\infty} f(x) e^{\lambda \int_1^{x} f(t)^2 dt} dx=0$$

where $\lambda$ is a complex number (with $0<\Re(\lambda)< \frac{1}{2}$) and where we impose following conditions on $f(x)$ near zero and infinity:

$$f(x)\sim x^{-\frac{1}{2}} \; \; \; ( x\to 0) $$

$$f(x)\sim x^{-\frac{1}{2} +a} \; \; \; ( x\to \infty) $$

(with $a>0$)

Any idea on the way to handle such a problem ?

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    $\begingroup$ You have a function to play around with and you just need two integrals to be zero (after separating into real and imaginary parts), so you'd expect gazillions of solutions for non-real $\lambda$. (The required asymptotics are not really stopping us, for example, you can take $f$ equal to these expressions near $0,\infty$.) $\endgroup$
    – Christian Remling
    Commented Dec 20, 2017 at 19:55
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    $\begingroup$ $=0$ identically in $\lambda$? Or $f$ may depend on $\lambda?$ $\endgroup$
    – Alexandre Eremenko
    Commented Dec 20, 2017 at 21:43
  • $\begingroup$ $f(x)$ does not depends on $\lambda$. $\endgroup$
    – Bertrand
    Commented Dec 21, 2017 at 21:02
  • $\begingroup$ @Christian The real and imaginary integrals are linked and $f(x)$ is a positive function so not obvious that we can have solutions... but I agree it seems we have many, then finding a demonstration is far more difficult. $\endgroup$
    – Bertrand
    Commented Dec 21, 2017 at 21:09

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