can we find a function $f(x)\not=0$,such that $\int_1^{\infty}\left(1-\frac{1}{x}\right)f(x)dx=0$
who can give an instance ?
thanks
can we find a function $f(x)\not=0$,such that $\int_1^{\infty}\left(1-\frac{1}{x}\right)f(x)dx=0$
who can give an instance ?
thanks
Let $g\in L^1([1,\infty))$ so that $ \int_1^\infty g(x)\,\mathrm{d}x=0 $ We can create such a $g$ by taking any $h_1,h_2\in L^1([1,\infty))$ and defining $ g(x)=h_1(x)\int_1^\infty h_2(t)\,\mathrm{d}t-h_2(x)\int_1^\infty h_1(t)\,\mathrm{d}t $
Now, define $f(x)=g(x-\log(x))$. Then, since $x-\log(x):[1,\infty)\mapsto[1,\infty)$ is bijective, $ \begin{align} \int_1^\infty\left(1-\frac1x\right)f(x)\,\mathrm{d}x &=\int_1^\infty f(x)\,\mathrm{d}(x-\log(x))\\[6pt] &=\int_1^\infty g(x-\log(x))\,\mathrm{d}(x-\log(x))\\[6pt] &=\int_1^\infty g(x)\,\mathrm{d}x\\[12pt] &=0 \end{align} $
Consider any $g(x)$ such that $\displaystyle \int_1^{\infty} g(x) dx = 0 $ and take $f(x) = \dfrac{g(x)}{1-1/x}$. One such class of function for $g(x)$ is $g(x) = \begin{cases} \alpha \dfrac{h(x)}{\displaystyle \int_1^a h(t) dt}; & 1 \leq x \leq a\\ -\alpha \dfrac{k(x)}{\displaystyle \int_a^{\infty} k(t) dt}; & a < x < \infty \end{cases}$ where $h(x)$ and $k(x)$ are such that $\left \vert \displaystyle \int_1^a h(t) dt \right \vert , \left \vert \displaystyle \int_a^{\infty} k(t) dt\right \vert < \infty$.
Consider the regular bounded functions $f_a$, $a\gt0$, defined by $ f_a:x\mapsto x\mathrm e^{-ax}. $ When $f=f_a$, the LHS is $1/(\mathrm e^{a}a^2)$. Choose $(a,b,c)$ positive such that $\frac1{\mathrm e^aa^2}=\frac{c}{\mathrm e^bb^2}, $ for example $a=1$, $b=2$ and $c=4\mathrm e$, then a solution is $f=f_a-cf_b$, for example, $ f(x)=x\cdot(\mathrm e^{-x}-4\mathrm e^{-2x+1}). $ More generally, for every $a\ne b$ positive, consider $ f(x)=x\cdot(a^2\mathrm e^{a(1-x)}-b^2\mathrm e^{b(1-x)}). $ Still more generally, for every nonzero function $g$ such that $\displaystyle\int_1^{+\infty}xg(x)\mathrm dx$ converges, consider, for every $a\ne b$ positive, $ f(x)=x\cdot(a^2g(a(x-1))-b^2g(b(x-1))). $