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I was wondering if we can use complex contour integration to evaluate the integral $$ \int_0^\infty \frac {\sin x}{1+x^3}dx. $$ Since the integrand is not even, we cannot extend the integration domain to $\mathbb R$ and use the upper semicircular contour $Re^{it}$, $0\le t\le \pi$. We cannot use the keyhole contour either, since the integral of $\frac{e^{iz}}{1+z^3}$ over the lower semicircle $Re^{it}$, $\pi\le t\le 2\pi$ does not converge to $0$ as $R\to \infty$.

Using Wolframalpha to evaluate the above improper integral, it does not give a closed form answer. Therefore I was wondering if complex integration is not applicable in this case.

pie
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Tongou Yang
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    It is unlikely to have any nice closed form, since replacing the $3$ with a $2$ has the same problem. You can obviously express it as a sum of values of trigonometric integrals via partial fraction decomposition, but I would be extremely hesitant to believe you'll get anything more (the antiderivative will also be expressible in terms of trig integrals, but will only be elementary for special values; I don't see any reason to suppose $0$ is special, since there is no symmetry about it (e.g. being odd/even), it's just a root) – Brevan Ellefsen Nov 26 '23 at 05:09
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    Mathematica gives $\frac{G_{1,7}^{5,1}\left(\frac{1}{46656}|\begin{array}{c} \frac{5}{6} \ \frac{1}{6},\frac{1}{3},\frac{1}{2},\frac{5}{6},\frac{5}{6},0,\frac{2}{3} \\end{array}\right)}{2 \sqrt{3 \pi }}$ were $G$ is the Meijer $G$ function. But I'm not sure I would call that better "closed form" than the original integral. (And the approximate numerical value is 0.610913.) – JimB Nov 26 '23 at 05:37
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    https://math.stackexchange.com/questions/2559331/improper-integral-integrating-over-contour-in-upper-half-plane looks closely related. Also https://math.stackexchange.com/questions/1057519/how-to-solve-int-0-infty-frac-cosaxx31dx – Gerry Myerson Nov 26 '23 at 05:59

1 Answers1

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Without complex integration.

Write $$1+x^3=(x+1)(x-a)(x-b)$$ where $$a=\frac{1+i \sqrt{3}}{2}\qquad \text{and} \qquad b=\frac{1-i \sqrt{3}}{2}$$ $$\frac 1{1+x^3}=\frac{1}{(a+1) (a-b) (x-a)}+\frac{1}{(b+1) (b-a) (x-b)}+$$ $$\frac{1}{(a+1) (b+1) (x+1)}$$ to face three integrals looking like $$I_k=\int \frac {\sin(x)}{x+k}\,dx$$ Let $x+k=t$, expand the sine to face the sine and cosine integrals. Back to $x$ $$I_k=\cos (k)\, \text{Si}(k+x)-\sin (k)\, \text{Ci}(k+x)$$

So, if $\Re(k)>0\lor k\notin \mathbb{R}$ $$J_k=\int_0^\infty \frac {\sin(x)}{x+k}\,dx=\text{Ci}(k) \sin (k)+\frac{1}{2} (\pi -2 \text{Si}(k)) \cos (k)$$

The formula is a bit too long to type here but you have the formula.

Claude Leibovici
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