Given $\int_{-\infty}^\infty f(x)dx=0,$ Prove/Disprove claims

Question

Prove/Disprove:

$(a)$ If $\int_{-\infty}^\infty f(x)dx=0,$ then there exists $x_0$ such that $\int_{x_0-x}^{x_0+x}f(t)dt=0$ for large enough $x$.

$(b)$ If $\int_{-\infty}^\infty f(x)dx=0,$ then there exists $x_0$ such that $\int_{x_0}^{\infty}f(t)dt=-\int_{-\infty}^{x_0}f(t)dt .$

My thoughts:

I think $(a)$ is true, I tried to split the integral:

$$\int_{-\infty}^\infty f(t)dt=\int_{-\infty}^{x_0-x}f(t)dt+\int_{x_0-x}^{x_0+x}f(t)dt+\int_{x_0+x}^\infty f(t)dt$$

and do some manipulations, but got stuck.

I think $(b)$ is false, I tried to construct a counter example of triangles with bases which get smaller as $x$ gets larger for $x>0$, and negative triangles with basis which gets bigger as $x$ gets smaller for $x<0,$ but not sure if that works.

Answer 1

In the comments of the question there were some concerns about wether the integrals exists or not. In this respect, note that, by definition, \begin{align}\int_{-\infty}^{+\infty} f(t) \,\mathrm{d}t &= \lim_{a\to -\infty}\lim_{b\to +\infty}\int_{a}^b f(t) \,\mathrm{d}t = \lim_{a\to -\infty}\lim_{b\to +\infty} \Big(\int_a^{x_0} f(t) \,\mathrm{d}t + \int_{x_0}^b f(t)\,\mathrm{d}t\Big) \\ &= \lim_{a\to -\infty}\int_a^{x_0} f(t) \,\mathrm{d}t +\lim_{b\to +\infty}\int_{x_0}^b f(t)\,\mathrm{d}t . \end{align} So the left-hand-side exists if and only if the right hand side exists. Therefore both $\lim\limits_{a\to -\infty}\int_a^{x_0} f(t)$ and $\lim\limits_{b\to +\infty}\int_{x_0}^b f(t)\,\mathrm{d}t$ exist. So we have \begin{align*} 0 = \int_{-\infty}^{+\infty} f(t) \,\mathrm{d}t = \int_{-\infty}^{x_0} f(t) \,\mathrm{d}t + \int_{x_0}^{+\infty} f(t) \,\mathrm{d}t \end{align*} By adding one integral on the otherside we get \begin{align} \int_{x_0}^{+\infty} f(t) \,\mathrm{d}t = - \int_{-\infty}^{x_0} f(t) \,\mathrm{d}t \end{align}

Answer 2

Part (a) is false. Just consider $$f(t)=\begin{cases}e^{-t}&t\geq0\\-\frac{1}{(t-1)^2}&t<0\end{cases}$$

Assume the claim is true for some $x_0$. Then take $x$ large enough so that $x_0-x$ is negative, $x_0+x$ is positive, and $\int_{x_0-x}^{x_0+x}f(t)\,dt=0$.

Then $$ \begin{align} 0=\int_{x_0-x}^{x_0+x}f(t)\,dt &=\int_{x_0-x}^{0}f(t)\,dt+\int_{0}^{x_0+x}f(t)\,dt\\ &=-1-\frac{1}{x_0-x-1}-e^{-(x_0+x)}+1\\ &=\frac{1}{x-x_0+1}-e^{-(x_0+x)}\\ \end{align}$$

and this is supposed to be true for all $x$ larger than some $M$. That is, for all large enough $x$, it is supposed to be true that $$e^{-(x_0+x)}=\frac{1}{x-x_0+1}$$ or rather that $$e^{-x}=\frac{e^{x_0}}{x-x_0+1}$$ which is true for at most one $x$-value not all "large enough" $x$.

It looks like part (b) has been answered by Nathanael.