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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.

Thanks for your attention.

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up vote 5 down vote accepted

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 $$\lim_{t\rightarrow 0^+}F(t)=\lim_{t\rightarrow 0^+}\int_0^{\infty} \frac{f(t y)}{1+y^2} dy= \int_0^{\infty} \lim_{t\rightarrow 0^+}\frac{f(t y)}{1+y^2} dy= \int_0^{\infty} \frac{f(0)}{1+y^2} dy=\frac{f(0)\pi}{2}.$$

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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
@Sun Usually this types of integrals are solved via such substitutions. At least, in a straight forward manner. However, both Pauls and sos's approaches are valid. Note that alhough it seems the opposite, sos's solution is more elementary than Paul's and yet it seems much complex. – Pedro Tamaroff Feb 23 '12 at 6:06

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 $$\begin{align*} \limsup_{t \to 0^+} \left| F(t) - \frac{\pi}{2} f(0) \right| = 0 & \Longrightarrow \lim_{t \to 0^+} \left| F(t) - \frac{\pi}{2} f(0) \right| = 0 \\ & \Longrightarrow \lim_{t \to 0^+} F(t) = \frac{\pi}{2} f(0) \end{align*}$$

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Thank you very much. :) – Sun Feb 18 '12 at 9:35

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