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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 ?


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Did you mean $f(x) \neq 0$ or $\forall x, f(x) \neq 0$? In the first case $f$ can be any function $\neq 0$ only on a countable set. Are there any other constraints on $f$, like continuity? – Karolis Juodelė Nov 24 '12 at 10:56
Where did this question originate? Is there any background? – robjohn Nov 24 '12 at 23:12
There is a exercise: Assume $\int_{1}^{+\infty}f(x)dx$ is convergence. then there exists $\xi \in (1,+\infty)$,such that $$\int_{1}^{+\infty}\frac{f(x)}{x}dx=\int_{1}^{\xi}f(x)dx$$. By second mean value theorem for integration ,for $\forall n \in \mathbb{N^+}$,there exists $\xi_n \in (1,n)$,such that$$\int_{1}^{n} \frac{f(x)}{x}dx=\int_{1}^{\xi_n}f(x)dx$$. when I let $n \rightarrow +\infty$, what about $\xi_n$, at least, If the claim of the exercise is right. I need to show $\varliminf_{n \rightarrow \infty}\xi_n \not= \infty$.so I need to add some constraints on $f(x)$. – Laura Nov 25 '12 at 3:31
except for $\int_1^{+\infty}f(x)dx$ is convergence, if there is no other constraints on $f(x)$. I want to construct a counter-example. so I post this post. – Laura Nov 25 '12 at 3:37
according to Karolis Juodelė's comment. I should to raise my questions more precisely. if $f(x) \not= 0$ only on a countable set. It is easy to find a lot of examples. – Laura Nov 25 '12 at 3:44
up vote 6 down vote accepted

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, $$ \begin{align} \int_1^\infty\left(1-\frac1x\right)f(x)\,\mathrm{d}x &=\int_1^\infty f(x)\,\mathrm{d}(x-\log(x))\\[6pt] &=\int_1^\infty g(x-\log(x))\,\mathrm{d}(x-\log(x))\\[6pt] &=\int_1^\infty g(x)\,\mathrm{d}x\\[12pt] &=0 \end{align} $$

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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 $\left \vert \displaystyle \int_1^a h(t) dt \right \vert , \left \vert \displaystyle \int_a^{\infty} k(t) dt\right \vert < \infty$.

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how is it immune to the f(1)=g(1)/0 case? – dineshdileep Nov 24 '12 at 11:46
@dineshdileep $f(x)$ only needs to be well defined on $\color{red}{(}1,\infty)$. – user17762 Nov 24 '12 at 18:38

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, $$ f(x)=x\cdot(a^2g(a(x-1))-b^2g(b(x-1))). $$

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