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I have no idea how to start, it looks like integration by parts won't work.

$$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$

If someone could shed some light on this I'd be very thankful.

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  • $\begingroup$ are you looking for exact value or just convergence? $\endgroup$ – Anurag A Jun 19 '14 at 9:37
  • $\begingroup$ With wolfram alpha I know that the answer is $\pi /8e$. I do not know how to come to a solution $\endgroup$ – user1043065 Jun 19 '14 at 9:41
  • $\begingroup$ Convergence is easy to prove since the function is bounded by an integrable function $\endgroup$ – user1043065 Jun 19 '14 at 9:44
  • $\begingroup$ I'm not so sure that it is bounded by an obvious integrable function. Perhaps I'm not thinking correctly. $\endgroup$ – Anurag A Jun 19 '14 at 9:45
  • $\begingroup$ M yields $\large{\pi \over 8{\rm e}} \approx 0.1445$. $\endgroup$ – Felix Marin Jun 20 '14 at 2:35
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It is well known that $\displaystyle\int_0^\infty\frac{\sin x}{x}\,dx$ converges. Then $$ \int_0^{\infty}\frac{x^5\sin x}{(1+x^2)^3}\,dx=\int_0^{\infty}\Bigl(\frac{x^5}{(1+x^2)^3}-\frac1x\Bigr)\sin x\,dx+\int_0^{\infty}\frac{\sin(x)}{x}. $$ This implies that the integral converges, since the the first integral on the right hand side converges absolutely,

To compute the value use calculus of residues. Let $$ f(z)=\frac{z^5\,e^{iz}}{(1+z^2)^3}. $$ $f$ is meromorphic on an open set contaning the upper half plane, with a pole of order $3$ at $z=i$. For $R>1$ let $C_R$ be the semicircle $z=R\,e^{it}$, $0\le t\le\pi$. Then $$ \int_{-R}^R\frac{x^5e^{ix}}{(1+x^2)^3}\,dx+\int_{C_R}f(z)\,dz=2\,\pi\,i\,\text{Res}(f;i). $$ Now you have to:

  1. Prove that $\lim_{R\to\infty}\int_{C_R}f(z)\,dz=0$
  2. Compute the residue.
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  • $\begingroup$ $f$ and $x^5\sin x/(1+x^2)^3 $ do not coincide on the reals. Is that no problem? $\endgroup$ – user1043065 Jun 19 '14 at 10:48
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    $\begingroup$ That the integral along the semi circle goes to 0 follows directl from the Jordan Lemma. $\endgroup$ – Zaid Alyafeai Jun 19 '14 at 12:01
  • $\begingroup$ I have used Lebesgue dominated convergence theorem for that, concluding that the function goes to zero $\lambda$ - almost everywhere. Would that also be correct? $\endgroup$ – user1043065 Jun 19 '14 at 12:05
  • $\begingroup$ I have edited the answer. The value of the integral will be the imaginary part of $2\,\pi\,i\,\text{Res}(f;i)$. $\endgroup$ – Julián Aguirre Jun 20 '14 at 11:51
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Consider the function $f(t)=e^{\large-\sqrt a|t|}$, then the Fourier transform of $f(t)$ is given by $$ \begin{align} F(x)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-ix t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-\sqrt a|t|}e^{-ix t}\,dt\\ &=\int_{-\infty}^{0}e^{\sqrt at}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-\sqrt at}e^{-ix t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(\sqrt a-ix)t}}{\sqrt a-ix} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(\sqrt a+ix)t}}{\sqrt a+ix} \right|_{0}^{t=v}\\ &=\frac{1}{\sqrt a-ix}+\frac{1}{\sqrt a+ix}\\ &=\frac{2\sqrt a}{x^2+a}. \end{align} $$ Next, the inverse Fourier transform of $F(x)$ is $$ \begin{align} f(t)=\mathcal{F}^{-1}[F(x)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(x)e^{ix t}\,dx\\ e^{-\sqrt a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sqrt a}{x^2+a}e^{ix t}\,dx\\ \frac{\pi e^{-\sqrt a|t|}}{\sqrt a}&=\int_{-\infty}^{\infty}\frac{e^{ix t}}{x^2+a}\,dx,\tag1 \end{align} $$ where $(1)$ can be rewritten as $$ \int_{0}^{\infty}\frac{e^{ix t}}{x^2+a}\,dx=\frac{\pi e^{-\sqrt at}}{2\sqrt a}.\tag2 $$ Now differentiate $(2)$ with respect to $a$ twice and with respect to $t$ five times, take the real part, and set $a=t=1$ yields \begin{align} \Re\left[\int_{0}^{\infty}\frac{\partial^2}{\partial a^2}\frac{\partial^5}{\partial t^5}\left(\frac{e^{ix t}}{x^2+a}\right)\,dx\right]_{t=1,\,a=1}&=\left.\frac{\partial^2}{\partial a^2}\frac{\partial^5}{\partial t^5}\left(\frac{\pi e^{-\sqrt at}}{2\sqrt a}\right)\right|_{t=1,\,a=1}\\ -2\int_{0}^{\infty}\frac{x^5\sin x}{(x^2+1)^3}\,dx&=-\frac{\pi}{4e}\\ \int_{0}^{\infty}\frac{x^5\sin x}{(x^2+1)^3}\,dx&=\large\color{blue}{\frac{\pi}{8e}}. \end{align}

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  • $\begingroup$ Forgot to mention that $a>0$ and $t>0$ in $(2)$. You may also refer to this link to obtain $(1)$ using double integral method (elementary way). $\endgroup$ – Tunk-Fey Jun 19 '14 at 11:59
  • $\begingroup$ Thank you very much for the effort! I am in a course of complex analysis so I think I'll use the other answer. I was still checking it so I did not yet check the answer acceptance button. sorry for that but still thanks for the effort! $\endgroup$ – user1043065 Jun 19 '14 at 12:04
  • $\begingroup$ @user1043065 It's OK, no hurt feeling from me. I just enjoy your problem, very nice. $\ddot\smile$ $\endgroup$ – Tunk-Fey Jun 19 '14 at 12:05
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An alternative to the obvious canonical answer given by Julián Aguirre: it is easy to compute $$\int_{-\infty}^{\infty} \frac{e^{i b x}}{x^2+a}dx $$ for $a, b > 0$, because the integrand has simple poles only. Now differentiate with respect to $a$ twice and with respect to $b$ 5 times, and set $a = b = 1$. The theorem you need to justify the differentiation can be found in your textbook.

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Note \begin{eqnarray} I&=&\int_0^\infty \frac{x[(x^2+1)-1]^2\sin x}{(1+x^2)^3}dx\\ &=&\int_0^\infty \frac{x\sin x}{1+x^2}dx-2\int_0^\infty \frac{x\sin x}{(1+x^2)^2}dx+\int_0^\infty \frac{x\sin x}{(1+x^2)^3}dx. \end{eqnarray} From $$ \int_0^\infty \frac{\cos(ax)}{b^2+x^2}dx=\frac{\pi}{2b}e^{-ab}, a>0, b>0, $$ we have \begin{eqnarray} \int_0^\infty \frac{x\sin(ax)}{b^2+x^2}dx&=&=-\frac{d}{da}\int_0^\infty \frac{\cos(ax)}{b^2+x^2}dt=\frac{\pi}{2e^{-ab}},\\ \int_0^\infty \frac{x\sin x}{(b^2+x^2)^2}dx&=&-\frac{1}{2b}\frac{d}{db}\int_0^\infty \frac{x\sin(ax)}{b^2+x^2}dt=\frac{a\pi}{4be^{-ab}},\\ \end{eqnarray} and \begin{eqnarray} \frac{d^2}{db^2}\int_0^\infty \frac{x\sin (ax)}{b^2+x^2}dx&=&8b^2\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^3}dx-2\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^2}dx. \end{eqnarray} The latter one implies \begin{eqnarray} \int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^3}dx=\frac{a(ab+1)\pi}{16b^3e^{-ab}}. \end{eqnarray} Thus $$ I=\frac{\pi}{8e}.$$

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