I am trying to solve the following integral using only complex analysis:

$$\int^{+\infty}_0\frac{x\sin x}{1+x^4}dx$$

So, as the function $f(x)$ inside the integral is even, the integral can be expressed this way:

$$\int^{+\infty}_0\frac{x\sin x}{1+x^4}dx = \frac{1}{2}\int^{+\infty}_{-\infty}\frac{x\sin x}{1+x^4}dx = I$$

Edit: If $x = Re (z)$, expressing $I$ in terms of $z$, I get:

$$ I = \int^{+\infty}_0\frac{ze^{iz}}{1+z^4}dz$$ $$f(z)= \frac{ze^{iz}}{1+z^4}$$

It has singularities on $z_0= \{\frac{1}{\sqrt{2}}(1+i),\frac{1}{\sqrt{2}}(1-i),\frac{1}{\sqrt{2}}(-1-i),\frac{1}{\sqrt{2}}(-1+i)\}$. If we take a contour $C$ with a upper-axis semi-circle B and the axis running from $−R→R$ and apply the residue theorem to the poles $z_0 = \{\frac{1}{\sqrt{2}}(1+i),\frac{1}{\sqrt{2}}(-1+i)\}$ I get


$$\int_C f(z)dz=2\pi i\sum_k Res(f,z_k)$$

Until this step, would this approach be correct?

  • 2
    $\begingroup$ The integral is real, but the point is that you have to define a corresponding complex function. Also, always replace trigonometric functions with exponential. So here we define $f(z)=\frac{ze^{iz}}{1+z^4}$. Now integrate it over a semi-circle in the upper half plane, and see what happens when you take the radius of the circle to infinity. $\endgroup$
    – Mark
    Commented May 10, 2021 at 15:11
  • $\begingroup$ Can you find the poles of $\frac{x \sin x}{1 + x^4}$? Reducing your interval of integration to $[0,N]$ or $[-N,N]$ (which we will let go to $\infty$ by a limit later) can you close this piece of a path into a closed path that includes one or more poles (so that the new path integral has a nonzero value)? Can you then show, in the limit $N \rightarrow \infty$, the added piece of path contributes zero (or some other constant)? Can you then write your integral in terms of (some of) the residues and the integral along the added path? $\endgroup$ Commented May 10, 2021 at 15:11
  • $\begingroup$ Thank you for your answers. Now, I've made some corrections in my question. Are they correct? $\endgroup$
    – user9867
    Commented May 10, 2021 at 16:05

1 Answer 1


You're on the right track, but some points are in order.

Firstly, I don't think you meant $x=\Re z$; all you really need to switch to your chosen complex integrand is $\sin z=\Im e^{iz}$.

Secondly, we need to check the arc $z=Re^{i\theta}$ from $\theta=0$ to $\theta=\pi$ makes a contribution that $\to0$ as $R\to\infty$. It's enough to note that if $|R|\ge\sqrt[4]{2}$ then $|1+R^4e^{4i\theta}|\ge R^4-1\ge\tfrac12R^4$ by the triangle inequality, and$$\left|\int_0^\pi\frac{Re^{i\theta}e^{iRe^{i\theta}}iRe^{i\theta}d\theta}{1+R^4e^{4i\theta}}\right|\le \int_0^\pi\left|\frac{Re^{i\theta}e^{iRe^{i\theta}}iRe^{i\theta}}{\tfrac12R^4}\right|d\theta=2R^{-2}\int_0^\pi e^{-R\sin\theta}d\theta\le2\pi R^{-2}.$$

Thirdly, the slickest way to do the residue calculations notes that if $p^4+1=0$ then$$\lim_{z\to p}\frac{z(z-p)}{1+z^4}=\lim_{z\to p}\frac{2z-p}{4z^3}=\frac{1}{4p^2}$$by L'Hôpital's rule, so the residue of $\frac{ze^{iz}}{1+z^4}$ at $p$ is $\frac{e^{ip}}{4p^2}$. I leave you to do the rest yourself. (Just remember we only care about poles of positive imaginary part, which halves your work.) You should find $I=\tfrac{\pi}{2}\sin\tfrac{1}{\sqrt{2}}\exp\tfrac{-1}{\sqrt{2}}$.

  • $\begingroup$ Thanks @J.G. But I don't understand why we need to check that the arc makes a contribution that tends to 0 as R tends to infinity. $\endgroup$
    – user9867
    Commented May 11, 2021 at 14:51
  • $\begingroup$ @conradDell Because the residue theorem computers the integral around the whole closed circuit; the line integral is that, minus the arc's integral. $\endgroup$
    – J.G.
    Commented May 11, 2021 at 15:06
  • $\begingroup$ So if the contribution tended to a constant instead of 0, would the final answer be the computed integral minus that constant? $\endgroup$
    – user9867
    Commented May 11, 2021 at 15:10
  • $\begingroup$ @conradDell That's what I said. $\endgroup$
    – J.G.
    Commented May 11, 2021 at 15:30
  • $\begingroup$ Thank you very much, @J.G. $\endgroup$
    – user9867
    Commented May 11, 2021 at 15:36

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