Prove $\int_{R^{+}} \frac{1}{1+x} a(x) dx > 0$ given $\int_{R^{+}} \frac{1}{x} a(x) dx > 0$ and $\int_{R^{+}} a(x) dx = 0$?

Question

For a function $a(x)$ that is defined on $x \in R^{+}$ and satisfies $\int_{R^{+}} a(x) dx = 0$, is it possible to prove that $\int_{R^{+}} \frac{1}{1+x} a(x) dx > 0$ if we know $\int_{R^{+}} \frac{1}{x} a(x) dx > 0$?

It seems that this inequality holds for some simple numerical examples, but I am wondering if it is always true. Thanks for helping me with this!

Answer 1

Let $b(s)=\left(\mathcal{L} a\right)(s)$. We have $$ \int_{\mathbb{R}^+}a(x)\,dx =\lim_{s\to 0^+} b(s)=0,\qquad \int_{\mathbb{R}^+}a(x)\frac{dx}{x}=\int_{\mathbb{R}^+}b(s)\,ds>0$$ $$ \int_{\mathbb{R}^+}a(x)\frac{dx}{x+1} = \int_{\mathbb{R}^+}e^{-s} b(s)\,ds $$ and $$b(s)=\frac{8}{(s+1)^2}-\frac{80}{(s+2)^2}+\frac{108}{(s+3)^2}$$

fulfills the first constraints, but $\int_{\mathbb{R}^+}e^{-s}b(s)\,ds$ is negative. It follows that $$ a(x) = 8x e^{-x} - 80x e^{-2x}+108 x e^{-3x} $$ is a counterexample to OP's conjecture.