Complex integration tricks for evaluating real integrals

Question

I was doing some homework and I stumbled upon this integral: $$ \int_{-\infty}^{\infty}dx \, \frac{\cos(x)-1+\frac{x^2}{2}}{x^4} $$ My first thoughts were:

1. at $\pm\infty$ it should converge, since the order of the denominator is bigger.
2. expanding in Taylor's series of the argument at $0$ should be: $\frac{1}{4!}+O(x^6)$. This means that the integral also converges near $0$, and the function is continuous (so there is an analytic continuation in $0$).

Im a little bit stuck because if the function is analytical its integral should be $0$? But this is not the case in the solution. Also, when adding two arcs (one at infinity and one around $0$) to close the loop, no other non-analytic points exists, so the residue theorem also says that the integral is $0$. What am I doing wrong? Thanks in advance for any help.

Answer 1

As stated in the comments, $\cos z$ grows exponentially unbounded on the circular arc from $R$ to $-R$ that traverses either half plane. The secret is to consider

$$I = \operatorname{Re}\left\{\lim_{R\to\infty}\int_{-R}^R\frac{e^{ix}-1-ix+\frac{1}{2}x^2+\frac{i}{6}x^3}{x^4}dx\right\}$$

The reason is that $e^{ix}$ does exponentially decay on the arc in the upper half plane whereas $\cos x$ did not, and the extra imaginary terms are to make sure the integrand is still analytic at $0$. For any complex integral you can add as many real/imaginary terms as you'd like if you're only taking the imaginary/real part of the integral at the end anyway, respectively.

Closing off the integral with the semicircular arc in the upper half plane, we do indeed get that there are no poles, hence

$$\int_{\Bbb{R}}\frac{e^{iz}-1-iz+\frac{1}{2}z^2+\frac{i}{6}z^3}{z^4}dz + \int_{\text{arc}}\frac{e^{iz}-1-iz+\frac{1}{2}z^2+\frac{i}{6}z^3}{z^4}dz = 0$$

We can compute the integral over the arc with the substitution $z = Re^{i\theta}$

$$= i\int_0^\pi \frac{\left(e^{iR\cos\theta}e^{-R\sin\theta}\right)}{R^3e^{3i\theta}}-\frac{\left(1+iRe^{i\theta}-\frac{R^2}{2}e^{2i\theta}\right)}{R^3e^{3i\theta}}+\frac{i}{6}\:d\theta$$

The relationship between the two integrals holds for all values of $R$, so simply take the limit as $R\to\infty$ at this stage

$$I = -\operatorname{Re}\left\{\lim_{R\to\infty}\int_{\text{arc}}\cdots\right\} = \int_0^\pi 0 - 0 + \frac{1}{6}\:d\theta = \boxed{\frac{\pi}{6}}$$

Answer 2

Alternatively, let $$f(z) = \frac{e^{iz}-1+ \frac{z^{2}}{2}}{z^{4}}.$$

About the origin, $f(z)$ has the Laurent series expansion $$ f(z)= \frac{i}{z^{3}}- \frac{i}{6z} + O(1). $$

If we integrate $f(z)$ around the indented semicircular contour $$[-R, -r] \cup re^{i[\pi, 0]} \cup [r, R] \cup Re^{i[0, \pi]}, $$ the integral vanishes on the large semicircle as $R \to \infty$ by a combination of Jordan's lemma and the estimation lemma (or by just the estimation lemma by itself).

And using the result from this question, the integral on the small clockwise-oriented semicircle goes to $- i \pi \left(- \frac{i}{6} \right) = -\frac{\pi}{6}$ as $r \to 0$.

Therefore, we have $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{ix}-1+ \frac{x^{2}}{2}}{x^{4}} \, \mathrm dx - \frac{\pi}{6} =0. $$

Equating the real parts on both sides of the above equation, we get $$\int_{-\infty}^{\infty} \frac{\cos(x)-1+ \frac{x^{2}}{2}}{x^{4}} \, \mathrm dx = \frac{\pi}{6}. $$

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In general, the Laurent series of $$g(z) = \frac{e^{iz}-\sum_{k=0}^n(-1)^k\frac{ x^{2k}}{(2k)!}}{z^{2(n+1)}} $$ about the origin is $$g(z) = \sum_{m=-n}^{0} \frac{i(-1)^{m+n} z^{2k-1}}{(2m+2n+1)!} + O(1).$$

Integrating $g(z)$ around the same contour, we get $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{ix}-\sum_{k=0}^n(-1)^k\frac{ x^{2k}}{(2k)!}}{x^{2(n+1)}}\, \mathrm dx - i \pi \left( \frac{i(-1)^{n}}{(2n+1)!} \right) =0. $$

Therefore, $$\int_{-\infty}^{\infty} \frac{\cos(x)-\sum_{k=0}^n(-1)^k\frac{ x^{2k}}{(2k)!}}{x^{2(n+1)}}\, \mathrm dx = \frac{(-1)^{n+1}\pi}{(2n+1)!}.$$

Answer 3

Consider $$I_n=\int_{-n\pi}^{+n \pi} \frac{\cos(x)-1+\frac{1}{2}x^2}{x^4}\,dx$$ where $n$ is an integer.

$$I_n=\frac{\text{Si}(n \pi )}{3}-\frac{3-(-1)^n}{3 \pi n}+\frac{2 \left(1-(-1)^n\right)}{3 \pi ^3 n^3}$$

$$\text{Si}(n \pi )=\frac \pi 2 -(-1)^n \left(\frac{1}{\pi n}-\frac{2}{\pi ^3 n^3}+O\left(\frac{1}{n^5}\right)\right)$$

$$I_n=\frac{\pi }{6}-\frac{1}{\pi n}+\frac{2}{3 \pi ^3 n^3}+O\left(\frac{1}{n^5}\right)$$ is a quite good approximation.

Edit

Using the generalization proposed by @Svyatoslav in comments

$$J_m=\int_{-n\pi}^{+n\pi} \frac{\cos x-\sum_{k=0}^m(-1)^k\frac{x^{2k}}{(2k)!}}{x^{2(m+1)}}dx$$ $$J_m=\frac{(-1)^{m+1}}{(2m+1)!}\left( \pi -\frac{2 (2 m+1)}{\pi n}+\frac{4 m (2 m-1) (2 m+1)}{3 \pi ^3 n^3}+O\left(\frac{1}{n^5}\right) \right)$$

Answer 4

INTEGRATION BY PARTS THRICE $$ \begin{aligned} \int_{-\infty}^{\infty} \frac{\cos x-1+\frac{x^2}{2}}{x^4} d x=&2 \int_0^{\infty} \frac{\cos x-1+\frac{x^2}{2}}{x^4} d x\\ = & -\frac{1}{3} \int_0^{\infty}\left(\cos x-1+\frac{x^2}{2}\right) d\left(\frac{1}{x^3}\right) \\ = & -\frac{2}{3}\left[\frac{\cos x-1+\frac{x^2}{2}}{x^3}\right]_0^{\infty}+\frac{2}{3} \int_0^{\infty} \frac{-\sin x+x}{x^3} d x \\ = & -\frac{1}{3} \int_0^{\infty}(-\sin x+x) d\left(\frac{1}{x^2}\right) \\ = & \frac{1}{3} \int_0^{\infty} \frac{-\cos x+1}{x^2} d x \\ = & -\frac{1}{3} \int_0^{\infty}(-\cos x+1) d\left(\frac{1}{x}\right) \\ = & \frac{1}{3} \int_0^{\infty} \frac{\sin x}{x} d x\\=&\frac{\pi}{6} \end{aligned} $$