In “Surely You’re Joking, Mr. Feynman!,” Nobel-prize winning Physicist Richard Feynman said that he challenged his colleagues to give him an integral that they could evaluate with only complex methods that he could not do with real methods:

One time I boasted, “I can do by other methods any integral anybody else needs contour integration to do.”

So Paul [Olum] puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration! He was always deflating me like that. He was a very smart fellow.

Does anyone happen to know what this integral was?

**Answer**

I doubt that we will ever know the exact integral that vexed Feynman.

Here is something similar to what he describes.

Suppose f(z) is an analytic function on the unit disk.

Then, by Cauchy’s integral formula,

∮γf(z)zdz=2πif(0),

where γ traces out the unit circle in a counterclockwise manner.

Let z=eiϕ.

Then

∫2π0f(eiϕ)dϕ=2πf(0).

Taking the real part of each side we find

∫2π0Re(f(eiϕ))dϕ=2πRe(f(0)).

(We could just as well take the imaginary part.)

Clearly we can build some terrible integrals by choosing f appropriately.

**Example 1.**

Let f(z)=exp2+z3+z.

This is a mild choice compared to what could be done …

In any case, f is analytic on the disk.

Applying (1), and after some manipulations of the integrand, we find

∫2π0exp(7+5cosϕ10+6cosϕ)cos(sinϕ10+6cosϕ)dϕ=2πe2/3.

**Example 2.**

Let f(z)=expexp2+z3+z.

Then

∫2π0exp(exp(7+5cosϕ10+6cosϕ)cos(sinϕ10+6cosϕ))×cos(exp(7+5cosϕ10+6cosϕ)sin(sinϕ10+6cosϕ))=2πee2/3.

**Attribution***Source : Link , Question Author : Argon , Answer Author : user26872*