Is it possible to evaluate $$\int_{0}^{\infty} \frac{e^{4ix}}{1+x^4} dx $$ without breaking the exponential part to the sine and cosine parts and if so by what contour?

It fails

[So far, I've made an attempt by using contour $[0,2R] \cup \gamma_R$ with positive orientation where $\gamma_R (t) = Re^{it} +R$.

Few calculations on the controlling part of the integral over the arc: $$ \int_{\gamma_R} f(z) dz = i \int_{0}^{\pi} \frac{\exp \left( 4i \{ Re^{it} + R \} \right)}{1+ (Re^{it} + R)^4}Re^{it} dt, $$ and by some calculations, we yield $$ \left| i \int_{0}^{\pi} \frac{\exp \left( 4i \{ Re^{it} + R \} \right)}{1+ (Re^{it} + R)^4}Re^{it} dt \right| \leq R \int_{0}^{\pi} \frac{\exp \left( -4R \sin t \right)}{\left| 1+ (Re^{it} + R)^4 \right| } dt. $$ And since $$ \left| 1+ (Re^{it} + R)^4 \right| \geq 1, $$ we get $$ R \int_{0}^{\pi} \frac{\exp \left( -4R \sin t \right)}{\left| 1+ (Re^{it} + R)^4 \right| } dt \leq R \int_{0}^{\pi} \exp \left( -4R \sin t \right) dt. $$ And by choosing $\delta > 0$ sufficiently small, on $ \delta \leq t \leq \pi - \delta$, we get $$ \exp \left( -4R \sin t \right) \leq \exp \left( -4R \sin \delta \right). $$ Thus $$ R \int_{0}^{\pi} \exp \left( -4R \sin t \right) dt \leq R \left( 2\delta + \frac{R}{\exp(R \sin \delta)} (\pi - 2\delta) \right) $$ ]

  • $\begingroup$ Can you find all the complex solutions of $z^4+1=0$? $\endgroup$ – Avitus Dec 11 '13 at 15:46
  • $\begingroup$ @Avitus Yes.... $\endgroup$ – jachilles Dec 11 '13 at 15:47
  • $\begingroup$ How did you get that the integral over $\gamma_R$ vanishes (in the limit for $R \to \infty$, presumably)? I don't think it does. $\endgroup$ – Daniel Fischer Dec 11 '13 at 16:36
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    $\begingroup$ No. In fact, you have $$\lim_{R\to\infty} \int_{\gamma_R} \frac{e^{4iz}}{1+z^4}\,dz = - \int_{0}^\infty \frac{e^{-4x}}{1+x^4}\,dx.$$ That much is easy to see. Unfortunately, I don't see an immediate easy way to compute the latter integral. $\endgroup$ – Daniel Fischer Dec 11 '13 at 17:00
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    $\begingroup$ No, note that there's no $i$ in the exponent of the last integral. If $R$ is large enough that the pole in the first quadrant is between $\gamma_R$ and the real axis, the integral over $\gamma_R$ is by Cauchy's integral theorem equal to the integral over $[2R,2R+ 2iR] + [2R+2iR, 2iR] + [2iR,0]$. The integrals over the first two segments vanish in the limit. (And I forgot a factor $i$ before the last integral, just to mention it.) $\endgroup$ – Daniel Fischer Dec 11 '13 at 17:13

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\int_{0}^{\infty}{\expo{4x\ic} \over x^{4} + 1}\,\dd x = \int_{-\infty}^{\infty}\Theta\pars{x}\,{\expo{4x\ic} \over x^{4} + 1}\,\dd x = \int_{-\infty}^{\infty}\pars{% \int_{-\infty}^{\infty}{\dd k \over 2\pi\ic}{\expo{\ic kx} \over k - \ic 0^{+}}}\, {\expo{4x\ic} \over x^{4} + 1}\,\dd x \\[3mm]&= \int_{-\infty}^{\infty}{\dd k \over 2\pi\ic}{1 \over k - \ic 0^{+}} \int_{-\infty}^{\infty}{\expo{\ic\pars{k + 4}x} \over x^{4} + 1}\,\dd x = \int_{-\infty}^{\infty}{\dd k \over 2\pi\ic}{1 \over k - \ic 0^{+}} \int_{-\infty}^{\infty}{\expo{\ic\verts{k + 4}x} \over x^{4} + 1}\,\dd x \\[3mm]&= \pp\int_{-\infty}^{\infty}{\dd k \over 2\pi\ic}{1 \over k}\, \underbrace{\int_{-\infty}^{\infty}{\expo{\ic\verts{k + 4}x} \over x^{4} + 1}\,\dd x} _{\ds{\equiv {\cal I}\pars{k}}} + \half\ \underbrace{\int_{-\infty}^{\infty}{\expo{\ic\verts{k + 4}x} \over x^{4} + 1}\,\dd x} _{\ds{{\cal I}\pars{0}}}\tag{1} \end{align}

The zeros of $x^{4} + 1 = 0$ are given by $x_{n} \equiv \expo{n\ic\pi/4}$. $n = 1, 3, 5, 7$ \begin{align} {\cal I}\pars{k}&\equiv \int_{-\infty}^{\infty}{\expo{\ic\verts{k + 4}x} \over x^{4} + 1}\,\dd x = 2\pi\ic\sum_{n = 1,3}\lim_{x \to x_{n}} \bracks{\pars{x - x_{n}}\,{\expo{\ic\verts{k + 4}x} \over x^{4} + 1}} = 2\pi\ic\sum_{n = 1,3}{\expo{\ic\verts{k + 4}x_{n}} \over 4x_{n}^{3}} \\[3mm]&= -\,{\pi\ic \over 2}\sum_{n = 1,3}x_{n}\expo{\ic\verts{k + 4}x_{n}} \\[3mm]&= -\,{\pi\ic \over 2}\braces{% {\root{2} \over 2}\pars{1 + \ic}\expo{\ic\verts{k + 4}\root{2}\pars{1 + \ic}/2} + {\root{2} \over 2}\pars{-1 + \ic}\expo{\ic\verts{k + 4}\root{2}\pars{-1 + \ic}/2}} \\[3mm]&= -\ic\,{\pi\root{2} \over 4}\,\expo{-\root{2}\verts{k + 4}/2}\braces{% \pars{1 + \ic}\expo{\ic\verts{k + 4}\root{2}/2} - \pars{1 - \ic}\expo{-\ic\verts{k + 4}\root{2}/2}} \\[3mm]&= -\ic\,{\pi\root{2} \over 4}\,\expo{-\root{2}\verts{k + 4}/2} 2\ic\Im\braces{\pars{1 + \ic}\expo{\ic\verts{k + 4}\root{2}/2}} \\[3mm]&= {\pi\root{2} \over 2}\,\expo{-\root{2}\verts{k + 4}/2}\bracks{% \sin\pars{{\root{2} \over 2}\,\verts{k + 4}} + \cos\pars{{\root{2} \over 2}\,\verts{k + 4}}} \end{align}

\begin{align} {\cal I}\pars{k} &= {\pi\root{2} \over 2}\bracks{\Im\phi\pars{k} + \Re\phi\pars{k}}\,,\qquad \phi\pars{k} \equiv \exp\pars{{\root{2} \over 2}\,\pars{-1 + \ic}\verts{k + 4}} \\[3mm] {\cal I}\pars{0} &= {\pi\root{2} \over 2}\expo{-2\root{2}}\bracks{% \sin\pars{2\root{2}} + \cos\pars{2\root{2}}} \end{align}

By replacing these results in $\pars{1}$, we get: \begin{align} \int_{0}^{\infty}{\expo{4x\ic} \over 1 + x^{4}}\,\dd x &= -\ic\,{\root{2} \over 4}\pp\int_{-\infty}^{\infty}{\dd k \over k}\bracks{% \Im\phi\pars{k} + \Re\phi\pars{k}} \\[3mm]&\phantom{=}+ {\pi\root{2} \over 4}\expo{-2\root{2}}\bracks{% \sin\pars{2\root{2}} + \cos\pars{2\root{2}}} \end{align}

Also \begin{align} &\pp\int_{-\infty}^{\infty}{\expo{-z\verts{k + 4}} \over k}\,\dd k \\[3mm]&=\expo{4z}\int_{-\infty}^{-4}{\expo{zk} \over k}\,\dd k + \expo{-4z}\lim_{\epsilon \to 0^{+}}\bracks{% \int_{-4}^{-\epsilon}{\expo{-zk} \over k}\,\dd k + \int_{\epsilon}^{\infty}{\expo{-zk} \over k}\,\dd k} \\[3mm]&=-\expo{4z}\int^{\infty}_{4}{\expo{-zk} \over k}\,\dd k + \expo{-4z}\int_{0}^{4}{\expo{-zk} - \expo{zk}\over k}\,\dd k + \expo{-4z}\int_{4}^{\infty}{\expo{-zk} \over k}\,\dd k \\[3mm]&= -2\sinh\pars{4z}\int_{4}^{\infty}{\expo{-zk} \over k}\,\dd k - \expo{-4z}\int_{0}^{4}{\sinh\pars{zk} \over k}\,\dd k \end{align} At this point the remaining integrals $\pars{~\mbox{with}\ z = {\root{2} \over 2}\bracks{1 - \ic}~}$ are quite simple.


Try to integrate $\int ln(z) \frac{e^{4iz}}{1+z^4} dz$ with the branch chosen to be from 0 to $\infty$ and the contour closed in the upper half plane. The path integral along the branch should give you what you want.

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    $\begingroup$ $\lvert e^{4iz}\rvert = e^{-4 \operatorname{Im} z}$. The part of the circle in the lower half plane doesn't play nicely. $\endgroup$ – Daniel Fischer Dec 11 '13 at 16:01
  • $\begingroup$ Guys do you think the contour $[0,2R] \cup \gamma_R$ with $\gamma_R(t) = Re^{it} + R$ would work? $\endgroup$ – jachilles Dec 11 '13 at 16:06
  • $\begingroup$ what if you just close the contour in the upper plane with the same branch choice? $\endgroup$ – Anode Dec 11 '13 at 16:12

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