Convergence of an Improper Integral

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SUMMARY

The discussion focuses on proving the convergence of the improper integral lim(σ→∞) ∫(1-x/σ)|f(x)|^2 dx, given that ∫ |f(t)|^2 dt is convergent for 0≤t<∞ and equals F. The key argument involves demonstrating that lim(σ→∞) ∫(x/σ)|f(x)|^2 dx converges to 0 by splitting the integral into two parts and applying the properties of square integrable functions. As σ approaches infinity, the first term approaches 0, while the second term remains bounded by an arbitrarily small ε, confirming the overall convergence to F.

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Let f(x) be a continuous functions on [0,∞) and that ∫ |f(t)|^2dt is convergent for 0≤t<∞.
Let ∫ |f(t)|^2dt for 0≤t<∞ equals F.
Show that lim(σ→∞) ∫(1-x/σ)|f(x)|^2 dx for0≤x≤σ converges to F.

I know that it needs to prove that lim(σ→∞) ∫(x/σ)|f(x)|^2 dx for0≤x≤σ converges to 0. Can anyone give me a hint??
 
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Since |f| is square integrable, then for any ε > 0, there exists a T such that the integral (T,∞) of |f|² < ∞.

For σ > T, split the integral of (x/σ)|f|² into 2 parts at T.

The integral from 0 to σ of (x/σ)|f|² < integral of (x/σ)|f|² from 0 to T + the integral of |f|² from T to ∞.

The first term -> 0 as σ -> ∞, while the second term < ε. Since ε is arbitrarily small, the integral -> 0.
 

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