I need help to prove the existence of a real function $h(x) \in C^1$ with condition that near zero $h(x) \sim \ln(x)$ and near infinity $\lim h(x)_{x \to \infty} = \infty$ such that following integrals are null (we have $0<a<\frac{1}{2}$):

$$\int_0^{\infty} h'(x)^{\frac{1}{2}} \; e^{(-a+ib) h(x)} dx =0 \;\; \, \;\; \,\;\; \, (1)$$

$$\int_0^{\infty} \big(h'(x)^{\frac{1}{2}} \; e^{(-a+ib) h(x)} -x^{-\frac{1}{2}-a+ib)}\big) x^{-\frac{1}{2}-a-ib)} dx =0 \;\; \, \;\; \,\;\; \, (2)$$

It is not difficult to show that an infinity of $h(x)$ satisfy (1): assuming $h(x)$ is strictly increasing we make the change of variable $y=h(x)$ and find that (1) becomes:

$$\int_{-\infty}^{\infty} G(y) \; e^{(-a+ib) y} dy =0 \;\; \, \;\; \,\;\; \, (3)$$

were $G$ is defined by $G(h(x))=h'(x)^{-\frac{1}{2}} $, so taking a $G$ function satisfying (3) (that we can easily construct by linear combination of different functions adjusted to have integral equal to zero) we deduce a function $h(x)$ exists as solution of the differential equation $G(h(x))=h'(x)^{-\frac{1}{2}} $.

But proving that some $h(x)$ function can satisfy (2) is far more difficult ! Even if commun sense tells us that such $h(x)$ functions exists. Any idea on how to treat this type of problem, any reference ? I am sure the same type of problem has already been treated but I was not able to find anything similar in literature.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.