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Where is the following function continuous

  1. Jan 24, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]f: [0,+\infty) \to \mathbb{R}: y \mapsto \int_0^{+\infty} y \arctan x \exp(-xy)\,dx.
    Show that this function is continuous in [tex]y[/tex] if [tex]y \neq 0[/tex]
    and discontinuous if [tex]y = 0[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I just can't get started, any hint?
  2. jcsd
  3. Jan 24, 2007 #2


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    Homework Helper

    Start by trying to simplify an expression for f(y+d)-f(y). Ultimately you want to show that for any epsilon>0, you can pick a delta so that |f(y+d)-f(y)|<epsilon for all d<delta.
  4. Jan 24, 2007 #3
    Here is my try:
    Choose a sequence [tex]y_n \in [0,+\infty )[/tex] such that [tex]y_n \to y (\neq 0)[/tex].
    Define the function [tex]g_n(x)=y_n \arctan x e^{-xy_n}[/tex], then its limit is [tex]g(x)=y\arctan x e^{-xy}[/tex].
    Note that [tex]|g_n(x)| \leq |y_n\arctan x|[/tex], it follows [tex]g_n[/tex] is integrable. Hence by dominated convergence thm we have
    [tex]\lim f(y_n)=\lim \int g_n \to \int g = f(y)[/tex].

    Am I right? Still no idea for the case [tex]y=0[/tex]
    Last edited: Jan 24, 2007
  5. Jan 25, 2007 #4


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    Homework Helper

    To use the dominated convergence theorem you need to find a function that bounds g_n for all n. In other words, this function can't have an n in it. Other than that you seem to be on the right track. The same idea should work for y=0.
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