How to prove: $\lim_{y\to\infty}\int_{1}^{\infty}\frac{f(x)}{x^2+y^2}dx=0 $

Question

Let $f:\left[0,\infty\right]\to \left[0,\infty\right]$ be locally integrable. It is integrable on every compact subinterval $I \subset \left[0,\infty\right]$, and assume that the improper integral:

$$\int_{1}^{\infty}\frac{f(x)}{x^2}\,dx$$ converges and is finite. Compute:

$$\lim_{y\to\infty}\int_{1}^{\infty}\frac{f(x)}{x^2+y^2}dx $$

The limit exists because the integral exists by comparison test for each $y$ and is decreasing function of $y$. I am sure that the answer is zero but I am not allowed to use the Dominated convergence theorem to solve this question, therefore I need either use $\epsilon$ $\delta$ definition or need to bound this integral and apply squeeze theorem somehow.

Answer 1

The idea is to split $\int_{1}^{\infty}\frac{f(x)}{x^2+y^2}dx$ into two parts: The integral over $[R, \infty)$ becomes small with large $R$ because $\int_{1}^{\infty}\frac{f(x)}{x^2}\,dx $ is finite, and then the integral over the bounded interval $[1, R]$ becomes small with large $y$.

For all $R > 0$ is $$ \begin{align} \int_{1}^{\infty}\frac{f(x)}{x^2+y^2}dx &= \int_{1}^{R}\frac{f(x)}{x^2+y^2}dx + \int_{R}^{\infty}\frac{f(x)}{x^2+y^2}dx \\ &\le \int_{1}^{R}\frac{f(x)}{y^2}\,dx + \int_{R}^{\infty}\frac{f(x)}{x^2}\,dx \\ &= \frac{1}{y^2} \int_1^R f(x) dx + \int_{R}^{\infty}\frac{f(x)}{x^2}\,dx \, . \end{align} $$

Given $\epsilon > 0$ we can first choose $R > 1 $ so large that $$ \int_{R}^{\infty}\frac{f(x)}{x^2}\,dx < \frac 12 \epsilon $$ and then $y_0 > 1$ so large that $$ \frac{1}{y_0^2} \int_1^R f(x) dx < \frac 12 \epsilon \, . $$ Then $$ \int_{1}^{\infty}\frac{f(x)}{x^2+y^2}dx < \frac 12 \epsilon + \frac 12 \epsilon = \epsilon $$ for all $y \ge y_0$.

Answer 2

As I suggested in comment, let's rewrite $\frac{f(x)}{x^2+y^2} = \frac{f(x)}{x^2} \frac{x^2}{x^2+y^2} = \phi(x)\psi(x,y)$. Using, that improper integral from $\phi$ exists and $\psi $ is monotone and bounded by $x$, then we can conclude uniform convergence of $\int\limits_{1}^{\infty}\frac{f(x)}{x^2+y^2}dx$. Now it is enough to enter the limit under the integral and get the answer.

Answer 3

Let $\displaystyle A=\int^\infty_{1}\frac{f}{x^2}dx$, since $f\ge 0$, we have $0\le A<\infty$

$$\forall \epsilon>0, \exists N_1>0, s.t. ~\forall n\ge N_1\Longrightarrow\int^\infty_{n}\frac{f}{x^2}dx<\frac\epsilon2$$

Take $N_2>0$, such that, $\displaystyle\frac A{N_2^2+1}<\frac\epsilon2$, next, let $N=\max(N_1, N_2)$, and when $y\ge N^2$, we have

$$\begin{align}\left|\int^\infty_{1}\frac{f}{x^2+y^2}dx-0\right|&=\int^N_{1}\frac{f}{x^2+y^2}dx+\int^\infty_N\frac{f}{x^2+y^2}dx\\ \\ &\le \int^N_{1}\frac{f}{x^2+N^4}dx+\int^\infty_N\frac{f}{x^2}dx\\ \\ &\le \int^N_{1}\frac{f}{x^2+x^2N^2}dx+\frac\epsilon2\\ \\ &\le \frac1{1+N^2}\int^\infty_{1}\frac{f}{x^2}dx+\frac\epsilon2\\ \\ &=\frac A{1+N^2}+\frac\epsilon2\\ \\ &<\epsilon \end{align}$$

Answer 4

Since $$\int_1^\infty \frac{f(x)}{x^2+y^2} \, {\rm d}x = \int_1^y \frac{f(x)}{x^2+y^2} \, {\rm d}x + \int_y^\infty \frac{f(x)}{x^2+y^2} \, {\rm d}x \, ,$$ it suffices to prove that $F(y)=\int\limits_1^y \frac{f(x)}{x^2+y^2} \, {\rm d}x$ vanishes as $y\rightarrow \infty$, as the second integral clearly does. If $F(y)>0$ would not vanish as $y\to\infty$, then the integral $\int\limits_1^\infty \frac{F(y)}{y} \, {\rm d}y$ diverges. However, $$\int_1^\infty \frac{{\rm d}y}{y} \int_1^y \frac{f(x)}{x^2+y^2} \, {\rm d}x=\int_1^\infty {\rm d}x \, f(x) \int_x^\infty \frac{{\rm d}y}{y(x^2+y^2)} = \frac{\log 2}{2}\int_1^\infty \frac{f(x)}{x^2} \, {\rm d}x < \infty$$ by Fubini's theorem.