Contour integration is a great tool for definite integrals, especially of the following type: $$\int_{-\infty}^\infty \frac{f(x)dx}{P(x)}$$ where $P(x)$ is a polynomial with no zeroes on the real axis (as far as I remember) and $f(x)$ is some appropriate regular function, which vanishes fast enough in the upper half-plane of $\mathbb{C}$. I'm not going to go into details here and refer everyone to a textbook definition.

Contour integration seems like magic to be, and I'm sure to most of the regular people who use it either during their education or for various applications like physics. It's simple for me to see how it works in the complex plane, but not how it applies to the real functions. Why would some poles which are not on the real line affect the value of a real integral which still can be seen as the area under the curve defined by the integrand?

I know that some integrals which can be solved by contour integration can be also solved in other ways, for example the most simple one:

$$I=\int_{-\infty}^\infty \frac{dx}{1+x^2}=\pi$$

Using $(1+i)(1-i)=1+x^2$ makes the contour integration very simple, however it's also simple to show that $I=4\int_0^1 \frac{dx}{1+x^2}$ and connect the following integral to the definition of arcangent and Lebniz series.

But not all of the integrals can be easily solved that way. What about:

$$\int_{-\infty}^\infty \frac{\cos x dx}{1+x^2}=\frac{\pi}{e}$$

Can this integral be solved by real methods? Are there real integrals with closed forms, which can only be evaluated by contour integration? And if they exist, then how can it be explained?

To be clear, I understand that we can numerically confirm the value to any precision by just using numerical methods for the real integral in question. Except for some highly oscillatory integrands (probably).

Yuriy S
  • 30,220
  • 5
  • 48
  • 168
  • 1
    The integral that stumped Feynman? :) –  Jan 22 '17 at 18:45
  • @OpenBall, there is a question on this site about it, but there is no explicit expression provided. And I'm interested in a more general answer. The question is here http://math.stackexchange.com/q/253910/269624 – Yuriy S Jan 22 '17 at 18:47
  • You address a question that, hopefully, has the following answer : "there are some integrals that can be computed by residue calculus and are not integrable by real methods". But it depends also of the category of "real methods" you are authorized to use. You should have a look at the method employed in CAS. Have a look for example to Risch algorithm and al. (http://mathematica.stackexchange.com/questions/6811/how-does-mathematica-integrate) – Jean Marie Jan 22 '17 at 19:05
  • @JeanMarie, I agree the definition of "real methods" is somewhat complicated since some of them a complex methods in disguise. But I hope to understand this on a more intuitive level – Yuriy S Jan 22 '17 at 19:14
  • If this is on intuitive grounds: fundamentally, in the complex residue integration case, an integral for example of type $\int_0^{\infty}f(x)dx$ does not necessitate to have a "computable" primitive function $F(x)$ – Jean Marie Jan 22 '17 at 19:21

1 Answers1


The answer to this question is likely no.

For the example provided here there are some answers using only real analysis:

Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis

What are different ways to compute $\int_{0}^{+\infty}\frac{\cos x}{a^2+x^2}dx$?

And so on.

My understanding is this: while most if not all of the integrals with a closed form can be evaluated by real methods, the solution would usually be very complicated and require some ad hoc tricks.

Complex analysis and contour integration actually give a unified framework for this problem and allow one to evlauate more and more complicated integrals using the same algorithms as for the simple ones.

Yuriy S
  • 30,220
  • 5
  • 48
  • 168