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How do I do this integral without using complex variable theorems? (i.e. residues)

$$\lim_{n\to \infty} \int_0^{\infty} \frac{\cos(nx)}{1+x^2} \, dx$$

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It's not about the integral, is it? It's about the limit. The point is finding the limit without computing the integral. This is a special case of the Riemann–Lebesgue lemma –  Harald Hanche-Olsen Jun 28 at 8:54
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Hint: If you integrate by part, you pick up a factor of $\frac{1}{n}$... –  achille hui Jun 28 at 8:56
    
One question: Why does not hold: $\lim_{n\to \infty} \int_0^{\infty} \frac{\cos(nx)}{1+x^2}dx$=$\int_0^{\infty} \frac{\cos(x)}{1+x^2}dx$. Actually the "n" does not really matter, doesnt it? –  Mesih Jun 28 at 16:45
    
I deleted my answer because I was not aware of your condition:"$\ldots$without using complex variable theorems$\ldots$". –  Felix Marin Jun 29 at 2:03
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3 Answers 3

up vote 3 down vote accepted

Let $t=nx$ then

$$\int_0^\infty\frac{\cos(nx)}{1+x^2}dx=\int_0^\infty\frac{n\cos t}{n^2+t^2}dt=\int_0^\infty f_n(t)\;dt$$ We have $$f_n(t)\xrightarrow{n\to\infty}0,\quad\forall t$$ and $$|f_n(t)|\le\frac{1}{1+t^2}\in L^1(0,\infty)$$ so we apply the dominated convergence theorem and the desired limit is $0$.

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@Analysis Sorry my answer isn't correct! –  Sami Ben Romdhane Jun 28 at 9:20
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This is funny. You said your answer is incorrect yet still got upvotes and accepted, but my answer got downvoted even if it is correct. LOL –  Tunk-Fey Jun 28 at 10:22
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Use integration by parts

$$\int_0^\infty \frac{\cos(nx)}{1+x^2} dx = \frac{1}{n} \int_0^\infty \frac{2\sin(nx)x}{(1+x^2)^2} dx$$

Now use the triangle inequality

$$\left|\int_0^\infty \frac{2\sin(nx)x}{(1+x^2)^2} dx\right|\le \int_0^\infty \frac{2x}{(1+x^2)^2} dx<\infty$$

So the limit is $0$.

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Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by $$ \begin{align} F(x)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-ix t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-ix t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-at}e^{-ix t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-ix)t}}{a-ix} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+ix)t}}{a+ix} \right|_{0}^{t=v}\\ &=\frac{1}{a-ix}+\frac{1}{a+ix}\\ &=\frac{2a}{x^2+a^2}. \end{align} $$ Next, the inverse Fourier transform of $F(x)$ is $$ \begin{align} f(t)=\mathcal{F}^{-1}[F(x)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(x)e^{ix t}\,dx\\ e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{x^2+a^2}e^{ix t}\,dx\\ \frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{ix t}}{x^2+a^2}\,dx\\ \frac{\pi e^{-a|t|}}{2a}&=\int_{0}^{\infty}\frac{e^{ix t}}{x^2+a^2}\,dx. \end{align} $$ Thus, taking the real part, putting $a=1$ and $t=n$, then $$ \Re\left[\int_{0}^{\infty}\frac{e^{inx}}{x^2+1}\,dx\right]=\int_{0}^{\infty}\frac{\cos nx}{x^2+1}\,dx=\large\color{blue}{\frac{\pi e^{-|n|}}{2}}. $$ Other method using double integral technique can be seen here. Consequently $$ \lim_{n\to\infty}\int_{0}^{\infty}\frac{\cos nx}{x^2+1}\,dx=\lim_{n\to\infty}\frac{\pi e^{-|n|}}{2}=\large\color{blue}0. $$

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