Uniform Convergence of an Improper Integral

Real Analysis
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Hamza Mahmood
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Joined: Fri Dec 04, 2015 4:54 pm

Uniform Convergence of an Improper Integral

#1

Post by Hamza Mahmood »

Suppose $\displaystyle f(x,\alpha )=\begin{cases}
\dfrac{e^{-\alpha x}-e^{-2\alpha x}}{x} & \text{ if } x\neq 0 \\\\
\alpha & \text{ if } x=0
\end{cases}$.
Is $\displaystyle \int_{0}^{\infty }f(x,\alpha )dx$ uniformly convergent for all $\alpha \geq 0$ ? If yes, then how? Please help.
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Grigorios Kostakos
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Re: Uniform Convergence of an Improper Integral

#2

Post by Grigorios Kostakos »

Hamza, Ι'm not so sure what you mean by
Hamza wrote:$...\displaystyle \int_{0}^{\infty }f(x,\alpha )\,dx$ uniformly convergent for all $\alpha \geq 0$ ?
but here is a solution:

It is appropriate to use Frullani's theorem which states: Let \(D=\bigl\{{(x,y)\in{\mathbb{R}}^2\;|\; x\geqslant0,\; a\leqslant y \leqslant b}\bigr\}\,, 0<a<b\,,\) and \(f(xy): D\longrightarrow{\mathbb{R}}\) is a function which is continuously differentiable on \(D\). If \(f(xy)\) takes finite values at \(x=0\) and \(x=\infty\) for all \(y\in[a,b]\), denoted as \(f(0)\) and \(f(\infty)\) respectively, then \[\displaystyle \int_{0}^{\infty}{\frac{f(ax)-f(bx)}{x}\, dx}=\bigl({f(0)-f(\infty)}\bigr)\log\tfrac{b}{a}\,.\] So, for the function \[\displaystyle g(x,\alpha )=\begin{cases}
\dfrac{{\rm{e}}^{-\alpha x}-{\rm{e}}^{-2\alpha x}}{x} & \text{ if } x\neq 0 \\\\
\alpha & \text{ if } x=0
\end{cases}\] and for $\alpha>0$, we have $f(xy)={\rm{e}}^{-xy}\,, \; x\in[0,+\infty)$ and
\begin{align*}
\displaystyle \int_{0}^{\infty }g(x,\alpha )\,dx&=\int_{0}^{\infty}{\frac{f(\alpha x)-f(2\alpha x)}{x}\, dx}\\
&=\bigl({f(0)-f(\infty)}\bigr)\log\tfrac{2\alpha}{\alpha}\\
&=(1-0)\,\log2\\
&=\log2\,.
\end{align*} On the other hand for $\alpha=0$, we have \begin{align*}
\displaystyle \int_{0}^{\infty }g(x,0)\,dx&=\int_{0}^{\infty}{\dfrac{1-1}{x}\, dx}=\\
&=\int_{0}^{\infty}0\, dx=\\
&=0\,.
\end{align*}

P.S.1. The integral $\int_{0}^{\infty }g(x,\alpha )\, dx$ does not depend on the value $g(0,\alpha )=\alpha$ at the single point $x=0$.
P.S.2. A proof of Frullani's theorem can be found here.
Grigorios Kostakos
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