QUESTION. It is looked for a closed solution for following real integrals

$\displaystyle\int_0^{\infty}\displaystyle\frac{\cos{ax}}{x^3+1}dx$ and $\displaystyle\int_0^{\infty}\displaystyle\frac{\sin{ax}}{x^3+1}dx$

while the constant $a$ can be of any value; not only $a = 0$ or $a \rightarrow \infty$.

To explain the meaning of $a$, the first integral is considered in a practical context. The integral can be used to describe the stress distribution in flanged beam near a loading point where $r$ describes the stiffness of the flange, $F$ is a single force and $t$ is the wall thickness of the web. For the stress at the web's top it holds:

$\sigma = \displaystyle\frac{F}{\pi t r} \displaystyle\int_0^{\infty}\displaystyle\frac{\cos{\frac{d}{r}x}}{x^3+1}dx$

enter image description here

For a distance far remote from the loading point (meaning $a = d/r = \infty$), the stress (as the integral) will amount zero. In contrast the stress (as the integral) is maximum for $a = d/r = 0$. In this case the $\cos$-function drops out. By contour integration one gets

$\displaystyle\int_0^{\infty}\displaystyle\frac{dx}{x^3+1} = \frac{2\pi\sqrt{3}}{9}$

To describe the stress distribution as a whole the knowledge of the integral for any value $a$ is required.

TRIED SOLUTION. I try to solve the aforementioned real integrals considering the following complex integral

$\displaystyle\oint_C f(z)\,dz = \displaystyle\oint_C\displaystyle\frac{e^{iaz}}{z^3+1}\cdot \log{z}\, dz= \displaystyle\oint_C\displaystyle\frac{\cos{az}}{z^3+1}\cdot \log{z}\, dz+ i \displaystyle\oint_C\displaystyle\frac{\sin{az}}{z^3+1}\cdot \log{z}\, dz$

I know that such an approach is appropriate for the integration of $\frac{1}{x^3+1}$ within the same limits.

By $\log{z}$ the considered complex-valued function $f(z)$ gets a branch point at $z = 0$. This point is connected with $+\infty$ to form a branch cut. The contour $C = C_1\cup C_2 \cup C_3 \cup C_4$ does not cross the branch cut (see picture).

intergation path

As the complex-valued function contains three poles $z_k$ inside $C$ it holds:

$\displaystyle\oint_C\displaystyle\frac{e^{iaz}}{z^3+1}\cdot \log{z}\,dz = 2\pi i \displaystyle\sum_{k = 1, 2,3}{\operatorname{Res} f(z): z_k}$

Looking at the paths it can be stated:

  • Path $C_2$: integral tends to zero for $R \rightarrow\infty$ (Jordan's Lemma)
  • Path $C_3$: integral tends to zero for $\varepsilon \rightarrow 0$ (can be proven)
  • Paths $C_1$ and $C_4$ would cancel out each other if the considered complex-valued function did not contain $\log{z}$. Through $\log{z}$ both paths differ by their imaginary parts.

    Path $C_1$: $\log{z} = \ln{r} + i \cdot 0$

    Path $C_4$: $\log{z} = \ln{r} + i \cdot 2\pi$

After extending the integrals for $C_1$ and $C_4$ I end up with:

$\displaystyle\oint_C\displaystyle\frac{e^{iaz}}{z^3+1} dz = - \displaystyle\sum_{k = 1, 2,3}{\operatorname{Res} f(z): z_k}$

I know for sure that the real integrals that I am looking for tend to zero for high values $a$ due to the oscillation, see picture. But my solution does not. I fear my approach is not correct. I would deeply appreciate any comment or hint.

graphs y=ƒ(x) for different values of parameter

REMARK #1: I want to add some information on parity for the sake of completeness as Lucian raised that issue although this approach did not turn out to be successful for my case.

I split the original function into an even and odd part.

$f(x) = f_{even}(x) + f_{odd}(x)$

$\displaystyle\frac{\cos{ax}}{x^3+1}= -\displaystyle\frac{\cos{ax}}{x^6-1}+\displaystyle\frac{x^3\cos{ax}}{x^6-1}$

This led to following integral:

$\displaystyle\int_0^{\infty}\displaystyle\frac{\cos{ax}}{x^3+1}dx = \displaystyle -\frac{1}{2}\int_{-\infty}^{+\infty}\displaystyle\frac{\cos{ax}}{x^6-1}dx + \displaystyle\int_{0}^{+\infty}\displaystyle\frac{x^3\cos{ax}}{x^6-1}dx$

The different integration limits can be attributed to the symmetrical properties of the even function (meaning that integral from $-\infty$ to 0 equals integral from 0 to $+\infty$) that makes things easier.

I did the same splitting for the complex-valued function and chose the two contours shown in the picture.

enter image description here

The integration of the even function along $C$ can be easily done as the integral of path $C_2$ dies out for $R \rightarrow\infty$ (Jordan's Lemma - here it holds indeed). For pole $z_1$ only half of the residuum has to be account for, as this pole is located on the boundary of the considered contour.

When integrating the odd part, the integral along path $C_2$ vanishes for $R \rightarrow\infty$ (similar to Jordan's Lemma), too. But the contribution of path $C_3$ cannot be neglected, unfortunately.

REMARK #2: Differentiation under the integral sign using the third derivative

I generalize the integral by introducing a parameter $b$:

$I(b) = \displaystyle\int_0^{\infty}\displaystyle\frac{\cos{ax}}{x^3+1}\left(\frac{x^3b^3}{6}+\frac{b^3}{6} +1\right)dx$

For $b = 0$ I get the integral that I am looking for.

Now I determine the third derivative with respect to $b$ hoping that it turns into an expression that can be easily integrated with respect to $x$. As $\displaystyle\frac{\cos{ax}}{x^3+1}$ is considered as a constant while deriving with respect to $b$ I end up with

$\displaystyle\frac{d^3I(b)}{db^3} = \displaystyle\frac{d^3}{db^3}\displaystyle\int_0^{\infty}\displaystyle\frac{\cos{ax}}{x^3+1}\left(\frac{x^3b^3}{6}+\frac{b^3}{6} +1\right)dx = \displaystyle\int_0^{\infty}\displaystyle\frac{\cos{ax}}{x^3+1} \frac{\partial^3}{\partial b^3}\left(\frac{x^3b^3}{6}+\frac{b^3}{6} +1\right)dx$

Auxiliary calculations:

  • $\displaystyle\frac{\partial}{\partial b}\left(\frac{x^3b^3}{6}+\frac{b^3}{6} +1\right) = \frac{x^3b^2}{2}+\frac{b^2}{2}$
  • $\displaystyle\frac{\partial^2}{\partial b^2}\left(\frac{x^3b^3}{6}+\frac{b^3}{6} +1\right) = x^3 b+b$
  • $\displaystyle\frac{\partial^3}{\partial b^3}\left(\frac{x^3b^3}{6}+\frac{b^3}{6} +1\right) = x^3 +1$

That gives a third derivative that does not converge.

$\displaystyle\frac{d^3I(b)}{db^3} = \displaystyle\int_0^{\infty}\cos{ax}\,dx = \left[\frac{1}{a}\sin{ax} \right]^{\infty}_0$

To conclude: although the used generalized integral does not seem to be appropriate, remark #2 regards the problem from a different angle that might be helpful for further discussions. Thanks to all contributors.

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  • why did you add the $\log z$ in your contour integral? – Ant Dec 08 '14 at 14:59
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    The log z gives me a branch point. Without log z path $C_1$ and $C_4$ would be identical. Their integrals would cancel out as the directions of both paths are opposite. – Layman33 Dec 08 '14 at 15:26
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    "Path $C_2$: integral tends to zero for $R \rightarrow\infty$ (Jordan's Lemma)" This is wrong, only the integral over half of $C_2$ - the upper half if $a > 0$, the lower half if $a < 0$ - tends to $0$ as $R\to\infty$ by Jordan's lemma. On the other half of $C_2$, the exponential $e^{iaz}$ blows up, $\lvert e^{iaz}\rvert = e^{-a\operatorname{Im} z}$, and that leads to the integral over that half of $C_2$ tending to a nonzero limit as $R\to \infty$. – Daniel Fischer Dec 08 '14 at 15:48
  • @Daniel Fischer: Thanks a lot. I was blind. Only for the case of $1/(x^3+1)$ my statement is right. – Layman33 Dec 08 '14 at 15:55
  • @Daniel Fischer: Can you imagine any other contour that makes sense? Or does my problem belong to the "unclosed cases of maths". – Layman33 Dec 08 '14 at 15:58
  • No inspiration yet. If I knew a contour (and integrand) that works, I'd have posted an answer rather than a comment. Maybe @RonGordon has an idea? – Daniel Fischer Dec 08 '14 at 16:03
  • The reason that $~\displaystyle\int_0^\infty\frac{\sin x}xdx~$ and $~\displaystyle\int_0^\infty\frac{\cos x}{1+x^2}dx~$ have such nice closed forms is due to the [parity](http://en.wikipedia.org/wiki/Even_and_odd_functions) of their integrand. Which is something that $\displaystyle\int_0^\infty\frac{\sin x}{1+x^2}dx,~$ for instance, along with the two integrals you posted, are lacking. – Lucian Dec 08 '14 at 18:42
  • @user_of_math: no it doesn't. – Ron Gordon Dec 08 '14 at 19:28
  • @Lucian: I added a remark dealing with parity ... might be interesting for you. – Layman33 Dec 09 '14 at 09:03
  • The problem is now you have singularities at $z=1$. This is by no means fatal; you may consider instead the principal values of the integrals. The problem is that in the odd integral, your contour won't. You may be better off to consider the complex form of the integral and taking the third derivative. – Ron Gordon Dec 09 '14 at 09:11
  • @Ron Gordon: I've messed it up. I mixed the pictures. When splitting the original function I get a new function with SIX poles instead of THREE. You are totally right!! – Layman33 Dec 09 '14 at 09:36
  • @Ron Gordon: By "taking the third derivative" you are talking on differentiating under the integral, don't you? – Layman33 Dec 09 '14 at 09:37
  • Yup. ${}{}{}{}{}{}{}{}{}{}$ – Ron Gordon Dec 09 '14 at 09:38
  • @Ron Gordon: Thanks for the hint. I will try my very best ... – Layman33 Dec 09 '14 at 09:42
  • @Ron Gordon: I added a remark #2 taking advantage of the third derivative. – Layman33 Dec 09 '14 at 13:38
  • The even part of your $($new$)$ integral does indeed possess a closed form expression. However, its odd part does not. That's because the sole purpose of the odd part is to cancel itself out when integrating over the entire real line. Zero is, after all, one of the most beautiful closed form expressions. – Lucian Dec 09 '14 at 17:03
  • Excuse me, I believe I may have spoken too soon: [This](http://i.stack.imgur.com/L0bhc.png) is the closed form of your integral's odd part, for $a=1$. – Lucian Dec 09 '14 at 17:15
  • @FelixMarin: $x$ is bound to the integral. How did it make it to the right side of the $=$? – robjohn Dec 10 '14 at 02:38
  • @robjohn Sorry. That was a typo. It's fixed. Thanks. $$ \int_{0}^{\infty}\cos\left(\, ax\,\right)\,{\rm d}x =\pi\delta\left(\, a\,\right) $$ in a 'distribution sense'. – Felix Marin Dec 10 '14 at 03:11

2 Answers2


The problem with the initial approach is that along $C_2$, $\dfrac{e^{iaz}}{x^3+1}$ does not vanish as $R\to\infty$. For $a\gt0$, it vanishes in the upper half-plane, but blows up exponentially in the lower half-plane.

Convert from Oscillating to Monotonic Integral

Let's compute the integral of $\dfrac{e^{iaz}}{z^3+1}$ over $\gamma$, where $$ \gamma=[0,R]\cup Re^{i[0,\pi/2]}\cup[iR,0]\tag{1} $$ The integral of $\dfrac{e^{iaz}}{z^3+1}$ along $Re^{i[0,\pi/2]}$ is less than $\frac{R\pi/2}{R^3-1}$ which vanishes as $R\to\infty$.

The residue of $\dfrac{e^{iaz}}{z^3+1}$ at $z=e^{i\pi/3}$ is $\frac13e^{-a\sqrt3/2}\left[\cos\left(\frac a2-\frac{2\pi}3\right)+i\sin\left(\frac a2-\frac{2\pi}3\right)\right]$.

Since $z=e^{i\pi/3}=\frac12+i\frac{\sqrt3}2$ is the only singularity inside $\gamma$, we get $$ \begin{align} &\frac{2\pi i}3e^{-a\sqrt3/2}\left[\cos\left(\frac a2-\frac{2\pi}3\right)+i\sin\left(\frac a2-\frac{2\pi}3\right)\right]\\ &=\int_\gamma\frac{e^{iaz}}{z^3+1}\mathrm{d}z\\ &=\int_0^\infty\frac{e^{iax}}{x^3+1}\,\mathrm{d}x-i\int_0^\infty\frac{e^{-ax}}{1-ix^3}\,\mathrm{d}x\\ &=\int_0^\infty\frac{e^{iax}}{x^3+1}\,\mathrm{d}x-i\int_0^\infty\frac{e^{-ax}(1+ix^3)}{1+x^6}\,\mathrm{d}x\tag{2} \end{align} $$ Solving for the integral we want: $$ \begin{align} \int_0^\infty\frac{e^{iax}}{x^3+1}\,\mathrm{d}x &=-\int_0^\infty\frac{x^3e^{-ax}}{1+x^6}\,\mathrm{d}x-\frac{2\pi}3e^{-a\sqrt3/2}\sin\left(\frac a2-\frac{2\pi}3\right)\\ &+i\left[\int_0^\infty\frac{e^{-ax}}{1+x^6}\,\mathrm{d}x+\frac{2\pi}3e^{-a\sqrt3/2}\cos\left(\frac a2-\frac{2\pi}3\right)\right]\tag{3} \end{align} $$ Therefore, separating real and imaginary parts, we have $$ \int_0^\infty\frac{\cos(ax)}{x^3+1}\mathrm{d}x =-\int_0^\infty\frac{x^3e^{-ax}}{1+x^6}\,\mathrm{d}x-\frac{2\pi}3e^{-a\sqrt3/2}\sin\left(\frac a2-\frac{2\pi}3\right)\tag{4} $$ and $$ \int_0^\infty\frac{\sin(ax)}{x^3+1}\mathrm{d}x =\int_0^\infty\frac{e^{-ax}}{1+x^6}\,\mathrm{d}x+\frac{2\pi}3e^{-a\sqrt3/2}\cos\left(\frac a2-\frac{2\pi}3\right)\tag{5} $$

Asymptotic Expansion

Equations $(4)$ and $(5)$ show that as $a\to\infty$, the integrals vanish. In fact, it is not difficult to obtain an asymptotic expansion from the right hand side of these integrals: $$ \int_0^\infty\frac{x^3e^{-ax}}{1+x^6}\,\mathrm{d}x \sim\frac{3!}{a^4}-\frac{9!}{a^{10}}+\frac{15!}{a^{16}}-\dots\tag{6} $$ and $$ \int_0^\infty\frac{e^{-ax}}{1+x^6}\,\mathrm{d}x \sim\frac{0!}{a^1}-\frac{6!}{a^7}+\frac{12!}{a^{13}}-\dots\tag{7} $$ Unfortunately, neither Mathematica nor I have found any simpler formulas for these exponential integrals. However, Mathematica does verify that equations $(4)$ and $(5)$ hold numerically for several values of $a$.

Convergent Series

Although equations $(4)$ and $(5)$ are valid for any $a\ge0$ and make the integral easier to compute numerically, and $(6)$ and $(7)$ give good approximations for large $a$, the question has now been amended to express interest in small $a$.

Integration by parts gives $$ \newcommand{\Ein}{\operatorname{Ein}} \begin{align} \int_a^\infty\frac{e^{-x}}{x}\mathrm{d}x &=\int_0^a\frac{1-e^{-x}}{x}\mathrm{d}x-\log(a)-\gamma\\[6pt] &=\Ein(a)-\log(a)-\gamma\tag{8} \end{align} $$ where $$ \begin{align} \Ein(x)&=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^k}{k\,k!}\tag{9}\\ \end{align} $$ or to remove oscillation for positive $x$: $$ \begin{align} e^x\Ein(x)&=\sum_{k=1}^\infty\frac{H_k}{k!}x^k\tag{10}\\ \end{align} $$ Therefore, $$ \begin{align} \int_0^\infty\frac{e^{-x}\mathrm{d}x}{x+a} &=e^a\int_a^\infty\frac{e^{-x}\mathrm{d}x}{x}\\[6pt] &=e^a\left(\Ein(a)-\log(a)-\gamma\right)\tag{11} \end{align} $$ A change of variables, partial fractions yields, and applications of $(10)$ and $(11)$ yield $$ \begin{align} &\int_0^\infty\frac{x^3e^{-ax}}{1+x^6}\,\mathrm{d}x\\ &=a^2\int_0^\infty\frac{x^3e^{-x}}{a^6+x^6}\,\mathrm{d}x\\ &=\small-\frac16\int_0^\infty\left(\frac1{x+ia}+\frac\omega{x+i\omega a}+\frac{\omega^2}{x+i\omega^2a}+\frac1{x-ia}+\frac\omega{x-i\omega a}+\frac{\omega^2}{x-i\omega^2a}\right)e^{-x}\,\mathrm{d}x\\ &=-\frac13\int_0^\infty\mathrm{Re}\left(\frac1{x+ia}+\frac\omega{x+i\omega a}+\frac{\omega^2}{x+i\omega^2a}\right)e^{-x}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\frac{H_{6k+2}}{(6k+2)!}a^{6k+2}\\ &+\small\frac13\left[\cos(a)-\cos\left(\tfrac12a\right)\cosh\left(\tfrac{\sqrt3}2a\right)-\sqrt3\sin\left(\tfrac12a\right)\sinh\left(\tfrac{\sqrt3}2a\right)\right](\log(a)+\gamma)\\ &+\frac\pi{18}\sin\left(\tfrac12a\right)\left[3\cosh\left(\tfrac{\sqrt3}2a\right)-2\sinh\left(\tfrac{\sqrt3}2a\right)\right]\\ &+\frac\pi{6\sqrt3}\cos\left(\tfrac12a\right)\left[2\cosh\left(\tfrac{\sqrt3}2a\right)-3\sinh\left(\tfrac{\sqrt3}2a\right)\right]\\ &-\frac\pi6\sin(a)\tag{12} \end{align} $$ and $(12)$ provides a real method to compute the integral appearing in $(4)$.

Similarly for the integral appearing in $(5)$ $$ \begin{align} &\int_0^\infty\frac{e^{-ax}}{1+x^6}\,\mathrm{d}x\\ &=a^5\int_0^\infty\frac{e^{-x}}{a^6+x^6}\,\mathrm{d}x\\ &=\small\frac16\int_0^\infty\left(\frac{i}{x+ia}+\frac{i\omega}{x+i\omega a}+\frac{i\omega^2}{x+i\omega^2a}-\frac{i}{x-ia}-\frac{i\omega}{x-i\omega a}-\frac{i\omega^2}{x-i\omega^2a}\right)e^{-x}\,\mathrm{d}x\\ &=-\frac13\int_0^\infty\mathrm{Im}\left(\frac1{x+ia}+\frac\omega{x+i\omega a}+\frac{\omega^2}{x+i\omega^2a}\right)e^{-x}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^{k+1}\frac{H_{6k+5}}{(6k+5)!}a^{6k+5}\\ &+\small\frac13\left[\sin(a)+\sin\left(\tfrac12a\right)\cosh\left(\tfrac{\sqrt3}2a\right)-\sqrt3\cos\left(\tfrac12a\right)\sinh\left(\tfrac{\sqrt3}2a\right)\right](\log(a)+\gamma)\\ &+\frac\pi{18}\cos\left(\tfrac12a\right)\left[3\cosh\left(\tfrac{\sqrt3}2a\right)-2\sinh\left(\tfrac{\sqrt3}2a\right)\right]\\ &-\frac\pi{6\sqrt3}\sin\left(\tfrac12a\right)\left[2\cosh\left(\tfrac{\sqrt3}2a\right)-3\sinh\left(\tfrac{\sqrt3}2a\right)\right]\\ &+\frac\pi6\cos(a)\tag{13} \end{align} $$

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  • The contour is identical with that of the odd function in the picture of the remark #1. Therefore, it struggles with path $C_3$, too. – Layman33 Dec 09 '14 at 14:29
  • @Layman33: actually, the integral along $C_3$ converges better than that along $C_1$ because of the exponential. I would call it a gift rather than a struggle. – robjohn Dec 09 '14 at 15:27
  • But how to unwrap this gift? – Layman33 Dec 09 '14 at 15:33
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    Do you have a reason to expect a closed form? If not, then at least this form makes it easier to compute an asymptotic expansion as $a\to\infty$. – robjohn Dec 09 '14 at 15:52
  • This is about as good as it gets for this one. (I derived an ugly form using cosine and sine integrals of complex argument...yuck.) – Ron Gordon Dec 09 '14 at 16:42
  • @robjohn: I added two pics showing the LEFT sides of Eq. (6) and (7). The pics are supposed to clarify that the integrals seem to be definite and of small size. When I develop the RIGHT side of Eq. (6) and (7) accounting for the first 100 elements I obtain huge values at least for $a$= 1. Am I wrong? – Layman33 Dec 09 '14 at 17:31
  • @Layman33: In general, asymptotic expansions are not convergent; they are not like Taylor expansions. These asymptotic expansions are meant for large values of $a$; the larger that $a$ is, the more terms you can use. If you take the first two terms of $(6)$, the error is $O\left(\frac1{a^{16}}\right)$. For example $$\int_0^\infty\frac{x^3e^{-100x}}{1+x^6}\mathrm{d}x=5.9999996371213076233\times10^{-8}$$ whereas $$\frac{3!}{100^4}-\frac{9!}{100^{10}}=5.9999996371200000000\times10^{-8}$$ The difference is about $\frac{15!}{100^{16}}=1.307674368\times10^{-20}$. – robjohn Dec 09 '14 at 18:47
  • @robjohn I clarified my question by giving a practical context. I hope this explains why I am concerned about the outcome of your solution letting $a = 1$. – Layman33 Dec 11 '14 at 14:50
  • @Layman33: I have added a method to compute the cosine integral using convergent series. I will add one for the sine integral in a while. – robjohn Dec 14 '14 at 21:01

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\cos\pars{ax} \over x^{3} + 1}\,\dd x} =a^{2}\int_{0}^{\infty}{\cos\pars{x} \over x^{3} + \verts{a}^{3}}\,\dd x =-\,{1 \over 3\verts{a}^{3}}\sum_{n\ =\ -1}^{1}x_{n}\int_{0}^{\infty} {\cos\pars{x} \over x - x_{n}}\,\dd x \\[5mm]&\mbox{where}\qquad x_{-1} \equiv \exp\pars{-\,{\pi\ic \over 3}}\verts{a}\,,\quad x_{0} \equiv -\verts{a}\,,\quad x_{1} \equiv \exp\pars{{\pi\ic \over 3}}\verts{a} \end{align}

Then, \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\cos\pars{ax} \over x^{3} + 1}\,\dd x} ={1 \over 3a^{2}}\dsc{\int_{0}^{\infty}{\cos\pars{x} \over x + \verts{a}}\,\dd x} -{2 \over 3\verts{a}^{3}}\,\Re\bracks{x_{1}\dsc{\int_{0}^{\infty} {\cos\pars{x} \over x - x_{1}}\,\dd x}}\quad\pars{1} \end{align}

However, \begin{align}&\dsc{\int_{0}^{\infty}{\cos\pars{x} \over x + \mu}\,\dd x} =\int_{\mu}^{\infty}{\cos\pars{x - \mu} \over x}\,\dd x \\[5mm]&=\cos\pars{\mu}\int_{\mu}^{\infty}{\cos\pars{x} \over x}\,\dd x +\sin\pars{\mu}\int_{\mu}^{\infty}{\sin\pars{x} \over x}\,\dd x \\[5mm]&=-\cos\pars{\mu}\,{\rm Ci}\pars{\mu} +\sin\pars{\mu}\bracks{{\pi \over 2} - \,{\rm Si}\pars{\mu}} \\[5mm]&=\dsc{{\pi \over 2}\,\sin\pars{\mu} -\cos\pars{\mu}\,{\rm Ci}\pars{\mu} -\sin\pars{\mu}\,{\rm Si}\pars{\mu}}\tag{2} \end{align} where $\ds{\,{\rm Ci}}$ and $\ds{\,{\rm Si}}$ are the Cosine Integral and the Sine Integral , respectively.

In using the result $\pars{2}$, the integral $\pars{1}$ can be expressed in terms of the Cosine Integrals and Sine Integrals.

Felix Marin
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  • This looks as if it leads to the answer Ron Gordon mentions in [a comment to my answer](http://math.stackexchange.com/questions/1057519/how-to-solve-displaystyle-int-0-infty-displaystyle-frac-cosaxx31dx#comment2156528_1059256). (+1) – robjohn Dec 10 '14 at 09:56
  • @robjohn Usually I read the other answers before I write my own one. However, I seldom read the comments to that answers. I just read the comment you mention right now. When I saw the $\tt cosine$ in the numerator and a polynomial in the denominator I got the relation to the $\,{\rm Ci}$ and ${\rm Si}$. I'm pretty sure $\tt\mbox{@Ron Gordon}$ and you have a lot of methods to handle many math-problems. Thanks – Felix Marin Dec 10 '14 at 16:15
  • Don't get me wrong, this is a perfectly good answer, and depending on whether they care about the use of the special functions $\operatorname{Ci}$ and $\operatorname{Si}$, this might be the answer the OP is looking for. However, it might be nice to write out the whole answer. – robjohn Dec 10 '14 at 16:38
  • @FelixMarin Thanks a lot for your contribution. Would you mind giving some more explanations how to get to your solution? – Layman33 Dec 11 '14 at 15:01
  • @robjohn I don't get you wrong. I find your comment as a pertinent one. Thanks. – Felix Marin Dec 13 '14 at 21:42