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How can I compute the following limit?

$$\lim_{a \to 0^+} \left(a \int_{1}^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\,\mathrm dx\right)$$

Any hints you can please give?

Cheers

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I end up with: $\displaystyle \lim_{a\to0+}\int_{a}^{\infty} e^{-y} cos\left( \frac{2\pi}{1+(y/a)^{2}}\right) dy$. Then I'm stuck. How do I apply DCT here? –  student Nov 6 '10 at 0:25
    
Well, we have that $\int_{a}^{\infty}=\int_{0}^{\infty}-\int_{0}^{a}=I_1-I_2$. Now $I_2$ goes to $0$ as $a\to 0$. In $I_1$ we should first multiply the numerator and denominator of the fraction under $\cos$ by $a^2$ and then pass to the limit using the DCT. –  Andrey Rekalo Nov 6 '10 at 0:45
    
You're almost there. The integrand is $\le1$, so $\int_0^a$ vanishes. Thus the limit is equal to $\lim\limits_{a\to0^+}\int_0^\infty e^{-y}\cos\left(\frac{2\pi a^2}{a^2+y^2}\right)\;\mathrm{d}y$. See Didier's answer. –  robjohn Feb 5 '12 at 14:49
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6 Answers

Hint: try a change of the independent variable $y:=ax$.

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This question is in the section of Lebesgue dominated convergence theorem, that's why I asked it here, any ideas using DCT? –  student Nov 5 '10 at 20:30
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If you make the substitution correctly, you will find that only the $\cos$ factor and the lower limit of integration depend on $a$. Then the dominated convergence theorem can be applied in (an almost) straightforward way. –  Andrey Rekalo Nov 5 '10 at 20:43
    
Is it valid to put the a inside the integral? –  student Nov 5 '10 at 20:53
2  
Sure. Integrals commute with multiplication by constants. –  Andrey Rekalo Nov 5 '10 at 20:59
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More generally, for every bounded measurable function $u$, consider $$ I_a(u)=a \int_0^{\infty} \mathrm e^{-ax}u(x)\mathrm dx,\qquad J_a(u)=a \int_1^{\infty} \mathrm e^{-ax}u(x)\mathrm dx. $$ You are interested in $\lim\limits_{a\to0^+}J_a(u)$ for $u(x)=\cos(2\pi/(1+x^2))$.

Since $I_a(u)-J_a(u)$ is $a$ times the integral on $(0,1)$ of a uniformly bounded function, when $a\to0^+$, $I_a(u)-J_a(u)\to0$. From now on, we study $I_a(u)$.

From here, several methods are available. The one I prefer is to note that $I_a(u)=\mathrm E(u(X_a))$ where $X_a$ is a random variable with exponential distribution of parameter $a$, hence $X_a$ is distributed as $X_1/a$. Since $X_1\gt0$ with full probability, $X_1/a\to+\infty$ with full probability. Thus, if $u$ has a limit $u^*$ at infinity, $u$ is bounded and $I_a(u)=\mathrm E(u(X_1/a))\to u^*$ when $a\to0^+$.

In your case, $u^*=\cos(0)=1$ hence $$ \lim_{a\to0^+}a \int_1^{\infty} \mathrm e^{-ax}\cos(2\pi/(1+x^2))\mathrm dx=1. $$

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Another (quick) idea. You can start by using the inequality $$ 1-\frac{t^2}{2} \leq \cos(t) \leq 1 + \frac{t^2}{2} $$ Then $$ e^{-ax} - e^{-ax}\frac{4\pi^2}{2(1+x^2)^2} \leq e^{−ax} \cos\left(\frac{2π}{1+x^2}\right) \leq e^{-ax} + e^{-ax}\frac{4\pi^2}{2(1+x^2)^2} $$ When you integrate this inequality, you will obtain $a\int_1^{+\infty} e^{-ax} dx$ (which is the limit of your integral) and $a\int_1^{+\infty} e^{-ax}\frac{4\pi^2}{2(1+x^2)^2} dx$ which tends towards $0$ since $$ 0 \leq a\int_1^{+\infty} e^{-ax}\frac{4\pi^2}{2(1+x^2)^2} dx\leq a\int_1^{+\infty} \frac{4\pi^2}{2(1+x^2)^2}dx$$

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} &\lim_{a \to 0^+}\bracks{a\int_{1}^{\infty}\expo{-ax} \cos\pars{2\pi \over 1 + x^{2}}\,\dd x} \\[3mm]&= \lim_{a \to 0^+}\braces{a\int_{1}^{\infty}\expo{-ax}\,\dd x - a\int_{1}^{\infty}\expo{-ax} \bracks{1 - \cos\pars{2\pi \over 1 + x^{2}}}\,\dd x} \\[3mm]&= \lim_{a \to 0^+}\braces{\int_{a}^{\infty}\expo{-x}\,\dd x - a\int_{1}^{\infty}\expo{-ax} \bracks{2\sin^{2}\pars{\pi \over 1 + x^{2}}}\,\dd x} \\[3mm]&= 1 - 2\quad\overbrace{% \lim_{a \to 0^{+}}\braces{a\int_{1}^{\infty}\expo{-ax} \sin^{2}\pars{\pi \over 1 + x^{2}}}\,\dd x}^{\ds{=\ 0\,,\quad \pars{~\mbox{see below}~}}}\tag{1} \end{align} $$\color{#0000ff}{\large% \lim_{a \to 0^+}\bracks{a\int_{1}^{\infty}\expo{-ax} \cos\pars{2\pi \over 1 + x^{2}}\,\dd x} = 1} $$ Notice that $\verts{\sin\pars{x}} \leq \verts{x}$ and $\expo{-ax} < 1$ when $ax > 0$. Then, $$ 0 \leq \verts{a\int_{1}^{\infty}\expo{-ax}\sin^{2}\pars{\pi \over 1 + x^{2}}\,\dd x} \leq \verts{a}\,\verts{\int_{1}^{\infty}{\pi^{2} \over \pars{x^{2} + 1}^{2}}\,\dd x} $$ such that the limit in the right hand side of $\pars{1}$ is, indeed, zero when $a \to 0^{+}$.

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First, what's $\lim_{a \to 0^+} \int_1^\infty ae^{-ax} \: dx$? Second, can you show that your integral is not so far from $\int_1^\infty ae^{-ax} \: dx$?

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This is my hint:

Because $a\int_{1}^{\infty} e^{-ax}\cos\left ( \frac{2\pi}{1+x^2} \right )dx=\lim_{x\to \infty}\left ( e^{-a}-e^{-ax} \right )+\int_{1}^{\infty}\frac{4\pi x e^{-ax}}{(1+x^2)^2}\sin\left ( \frac{2\pi}{1+x^2} \right )dx$

Hence, the limit does not exist!

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because not exist $\lim_{\begin{matrix}a\to 0^+\\x\to \infty\end{matrix}}e^{-ax}$ –  Iloveyou Nov 25 '13 at 15:21
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