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

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.

• Сan you use the uniform convergence of the integral? Jun 6 at 11:08

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 $$y \ge y_0$$.

• Thank you so much. I put a lot of effort but I could not recognize $\frac{f(x){x^{2}+y^{2}<\frac{f(x)}{y^2}$ it is clear now after using this. Jun 6 at 11:33

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.

• The upper limit of this integral is infinity, the uniform convergence has no help to switch the limit and integration. Jun 6 at 11:59
• @MathFail. Which integral has upper limit infinity? Jun 6 at 12:58
• OP's integral, from $1$ to infinity, even though it is uniformly convergent, you can't switch limit and integration. Jun 6 at 13:07
• This is corrrect. If the improper integral is uniformly convergent as shown here with Abel's test, then you can switch the limit and integral. When $F_n(y) = \int_1^n g(x,y) \, dx$ converges uniformly to $F(y) = \int_1^\infty g(x,y) \, dx$, then continuity of $F_n$ implies continuity of $F$ and $\lim_{n \to \infty}\lim_{y \to y_0}F_n(y) = \lim_{y \to y_0}\lim_{n \to \infty}F_n(y)= F(y_0)$. This can be extended for limits as $y \to \infty$.
– RRL
Jun 6 at 13:20
• See The Elements of Real Analysis by Bartle.
– RRL
Jun 6 at 13:21

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}

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.