I’ve found here the following integral.

I=∫10sin(π(1−x))xx(1−x)1−xdx=∫10sin(πx)xx(1−x)1−xdx=πe24

I’ve never seen it before and I also didn’t find the evaluation on math.se. How could we verify it?

If it is a well-known integral, then could you give a reference?

**Answer**

This one can be done with “residue at infinity” calculation. This method is shown in the Example VI of http://en.wikipedia.org/wiki/Methods_of_contour_integration .

First, we use zz=exp(zlogz) where logz is defined for −π≤argz<π.

For (1−z)1−z=exp((1−z)log(1−z)), we use log(1−z) defined for 0≤arg(1−z)<2π.

Then, let f(z)=exp(iπz+zlogz+(1−z)log(1−z)).

As shown in the Ex VI in the wikipedia link, we can prove that f is continuous on (−∞,0) and (1,∞), so that the cut of f(z) is [0,1].

We use the contour: (consisted of upper segment: slightly above [0,1], lower segment: slightly below [0,1], circle of small radius enclosing 0, and circle of small radius enclosing 1, that looks like a dumbbell having knobs at 0 and 1, can someone edit this and include a picture of it please? In fact, this is also the same contour as in Ex VI, with different endpoints.)

On the upper segment, the function f gives, for 0≤r≤1,

exp(iπr)rr(1−r)1−rexp((1−r)2πi).

On the lower segment, the function f gives, for 0≤r≤1,

exp(iπr)rr(1−r)1−r.

Since the functions are bounded, the integrals over circles vanishes when the radius tend to zero.

Thus, the integral of f(z) over the contour, is the integral over the upper and lower segments, which contribute to

∫10exp(iπr)rr(1−r)1−rdr−∫10exp(−iπr)rr(1−r)1−rdr

which is

2i∫10sin(πr)rr(1−r)1−rdr.

By the Cauchy residue theorem, the integral over the contour is

−2πiResz=∞f(z)=2πiResz=01z2f(1z).

From a long and tedious calculation of residue, it turns out that the value on the right is

2iπe24.

Then we have the result:

∫10sin(πr)rr(1−r)1−rdr=πe24.

**Attribution***Source : Link , Question Author : user153012 , Answer Author : Sungjin Kim*