I would like to know if we can find a real function $v(x)$ and a complex function $f(x)$, such that they solve the following differential equation (with $\alpha$ a complex, $0<Re(\alpha)<1$):

$$(x^2 f(x)')'+v(x)f(x) = \alpha f(x)$$

with the constraint that the imaginary part of the following integral is zero, i.e.,

$$Im (\int_0^{\infty} \ln(x) \Big(\mathcal{F}_c \circ\phi (v(x) f(x))\Big) \Big(\mathcal{F}_c \circ\phi (\overline{f(x)})\Big) dx)=0 $$

denoting $\phi(g(x))=\frac{1}{x} g(\frac{1}{x})$ and $\mathcal{F}_c$ to be the cosine transform $\mathcal{F}_c (f(x))=\int_0^{\infty} f(t) \cos(xt) dt $.

If there was no $\ln(x)$ in the integral above, the result would be direct as all couples $(v(x),f(x))$ that are solutions of the differential equation would verify the second condition (using a property of cosine transform, providing the terms are well defined). However, due to the term $\ln(x)$ in the integral I did not manage to find the couples $(v(x),f(x))$ which fulfill the conditions above.

Any idea on a way to treat problems of this type?

if we can find$v(x)$ a real function and $f(x)$ a complex function-- ### --I did not managed tocharacterisethe couples (v(x),f(x)) fullfilling the conditions above--. So, which verb is the correct one? $\endgroup$ – fedja Dec 18 '17 at 5:30