# Find $\lim\limits_{t\to 0^+} \int_0^{\infty} \frac{t f(x)}{t^2+x^2} dx$

The question is:

Let $f(x)$ be bounded and continuous on $[0,\infty)$. Let $\displaystyle F(t)=\int_0^{\infty} \frac{t f(x)}{t^2+x^2} dx$ for $t>0$.

Find $\displaystyle \lim_{t\to 0^+} F(t)$.

If I set $|f(x)|\leq M$, then I can obtain $|F(t)|\leq \frac{\pi}{2} M$. But I can not find the limit.

I try to rewrite $\displaystyle F(t)=\int_0^{\infty} \frac{f(t y)}{1+y^2} dy$.

I think if I can put the limit into the integration, then $$\displaystyle \lim_{t\to o^+} F(t)=\int_0^{\infty} \frac{\lim\limits_{t\to 0^+}f(t y)}{1+y^2} dy=\frac{\pi}{2} f(0).$$

But I don't know whether I can do. I hope I can receive some hints or method in here.

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As you have done, by using the substitution $x=ty$, we can rewrite $F(t)$ as $\displaystyle F(t)=\int_0^{\infty} \frac{f(t y)}{1+y^2} dy$. By assumption, $f$ is bounded, i.e. $|f(ty)|\leq M$ for some constant $M$. This implies that $$\left|\frac{f(t y)}{1+y^2}\right|\leq \frac{M}{1+y^2}\mbox{ for all }y\in[0,\infty).$$ Note that the function $\displaystyle\frac{M}{1+y^2}$ is integrable on $[0,\infty)$, for $$\int_0^\infty \frac{M}{1+y^2}dy=\frac{M\pi}{2}<\infty.$$ By Dominated convergence theorem, we have $$\lim_{t\rightarrow 0^+}F(t)=\lim_{t\rightarrow 0^+}\int_0^{\infty} \frac{f(t y)}{1+y^2} dy= \int_0^{\infty} \lim_{t\rightarrow 0^+}\frac{f(t y)}{1+y^2} dy= \int_0^{\infty} \frac{f(0)}{1+y^2} dy=\frac{f(0)\pi}{2}.$$
Thank you very much. I think, the key is that I used the substitution $x=ty$. In fact, that's not easy to think for me. Is there another way if without the substitution? –  Sun Feb 18 '12 at 9:34
Split integral into two parts. Let $\epsilon > 0$ and let $\delta > 0$ such that $|x| \leq \delta$ implies $|f(x) - f(0)| \leq \epsilon$. Then \begin{align*} \left| F(t) - \frac{\pi}{2} f(0) \right| & = \left| \int_{0}^{\infty} \frac{f(t x) - f(0)}{1+x^2} \; dx \right| \\ & \leq \int_{0}^{\infty} \frac{|f(t x) - f(0)|}{1+x^2} \; dx \\ & = \int_{0}^{\delta / t} \frac{|f(t x) - f(0)|}{1+x^2} \; dx + \int_{\delta / t}^{\infty} \frac{|f(t x) - f(0)|}{1+x^2} \; dx \\ & \leq \epsilon \int_{0}^{\delta / t} \frac{dx}{1+x^2} + \int_{\delta / t}^{\infty} \frac{2M}{1+x^2} \; dx \\ & \leq \frac{\pi}{2} \epsilon + \int_{\delta / t}^{\infty} \frac{2M}{1+x^2} \; dx. \end{align*} Thus taking $\limsup_{t \to 0^+}$, we obtain $$\limsup_{t \to 0^+} \left| F(t) - \frac{\pi}{2} f(0) \right| \leq \frac{\pi}{2}\epsilon.$$ Since this is true for any $\epsilon > 0$, we must have \begin{align*} \limsup_{t \to 0^+} \left| F(t) - \frac{\pi}{2} f(0) \right| = 0 & \Longrightarrow \lim_{t \to 0^+} \left| F(t) - \frac{\pi}{2} f(0) \right| = 0 \\ & \Longrightarrow \lim_{t \to 0^+} F(t) = \frac{\pi}{2} f(0) \end{align*}