Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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

share|improve this answer
    
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
1  
@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
add comment

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

share|improve this answer
    
Thank you very much. :) –  Sun Feb 18 '12 at 9:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.