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Dec 21, 2017 at 21:09 comment added Bertrand @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.
Dec 21, 2017 at 21:02 comment added Bertrand $f(x)$ does not depends on $\lambda$.
Dec 20, 2017 at 21:43 comment added Alexandre Eremenko $=0$ identically in $\lambda$? Or $f$ may depend on $\lambda?$
Dec 20, 2017 at 19:55 comment added Christian Remling 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 $0,\infty$.)
Dec 20, 2017 at 18:22 history asked Bertrand CC BY-SA 3.0