∫dx x^2 /(1+ exp(sinx)) {for x from –pi/2 to +pi/2}; repost?

Question

rumor spread this integral can be also evaluated by means of complex numbers; please show work as I’m quite unaware how to do it this way;

No tsu, it is 1.292 by numeric means, please revise your answer!

Sin(x) is not = exp{(-i/2)*(e^x - e^(-x))};
Sin(x) = (1/2i)*(e^(ix) - e^(-ix)) = -i*sh(ix);
please tsu revise your answer; it looks very scientifically, but it�s wrong from the start! Do it immaculately!

Answer 1

The method I have so far is very similar to the method you already know, just using complex variables.

Consider the directed (CCW) closed contour defined by
C1: z = π/2 * exp(iθ), 0 < θ < π
C2: z = π/2 * exp(iθ), π < θ < 2π
or C2: z = -π/2 * exp(iθ), 0 < θ < π

C1+C2: circle with radius π/2 about the origin.

f(z) = z^2 / [1 + exp(sinz)]
Since f(z) has no poles inside or on C1+C2,
∫ {C1+C2} f(z) dz = 0
Note that f(z) does have poles, but outside the region bounded by C1+C2.

In parametric form:
C1: dz = i*π/2 exp(iθ) dθ
C2: dz = -i*π/2 exp(iθ) dθ

So ∫{C1} f(z) dz =
∫{0,π} (π/2)^2 exp(2iθ)* i(π/2) exp(iθ) dθ / [1 + exp{sin(π/2 exp(iθ))}]
= ∫{0,π} i*(π/2)^3 exp(3iθ) dθ/ [1 + exp(Y)] ...(1)

Similarly:
∫{C2} f(z) dz
= -∫{0,π} i*(π/2)^3 exp(3iθ) dθ/ [1 + exp(-Y)] ...(2)

Now let C be the line segment [-π/2, π/2]
Using similar reasoning as above
∫{C1+C} f(z) dz = 0
and ∫{-C+C2} f(z) dz = 0

Subtracting the two:
∫{C1-C2+2C} f(z) dz = 0
or ∫{C} f(z) dz = 1/2 ∫{C2-C1} f(z) dz ...(3)
The LHS of (3) is what we're interested in finding.

From (1) and (2), the RHS of (3) works out to be:
-1/2 ∫{0,π} i*(π/2)^3 exp(3iθ) dθ* [1/{1+exp(-Y)} + 1/{1+exp(Y)}]

1/{1+exp(-Y)} + 1/{1+exp(Y)} = 1

So RHS(3) = -1/2 ∫{0,π} i*(π/2)^3 exp(3iθ) dθ
= -1/2 *i/3i * (π/2)^3 * exp(3iθ)| {0, π}
= -1/2 * 1/3 * (π/2)^3 * (-1 - 1)
= π^3 / 24 = LHS(3)

Ans. π^3 /24

Answer 2

Let x = -t => dx = - dt
I = ∫dx x^2 /(1+ exp(sinx)) {for x from –π/2 to +π/2}; ... ( 1 )
I = - ∫dt t^2*e(sint) /(1+ exp(sint)) {for t from π/2 to -π/2};
I = ∫dt t^2*e(sint) /(1+ exp(sint)) {for t from -π/2 to π/2};
I = ∫dx x^2*e(sinx) /(1+ exp(sinx)) {for x from -π/2 to π/2}; ... ( 2 )
Adding ( 1 ) and ( 2 ),
2I = ∫x^2 dx {for x from -π/2 to π/2};
I = (1/2) [x^3/3] {for x from -π/2 to π/2};
I = (1/2) [ (π)^3/24 + (π)^3/24 ]
I = (π)^3 / 24 = 1.292.

Complex Number Method:

I = ∫dx x^2 /(1+ exp[-i*sh(ix)]) {for x from –π/2 to +π/2};
I = ∫dx x^2*e^[i*sh(ix)]) / (1+ exp[i*sh(ix)]) ... ( 1 )
{for x from –π/2 to +π/2};

Let x = -t => dx = -dt {for t from π/2 to -π/2};
I = -∫dt t^2*e^[-i*sh(it)]) / (1+ exp[-i*sh(it)]) {for t from π/2 to -π/2};
I = ∫dt t^2*e^[-i*sh(it)]) / (1+ exp[-i*sh(it)]) {for t from -π/2 to π/2};
I = ∫dx x^2*e^[-i*sh(ix)]) / (1+ exp[-i*sh(ix)]) {for x from -π/2 to π/2};
I = ∫dx x^2* / (1+ exp[i*sh(ix)]) ... ( 2 )
{for x from -π/2 to π/2};

Adding ( 1 ) and ( 2 ),
2I = ∫dx x^2 {for x from -π/2 to π/2};
I = (1/2) [x^3/3] {for x from -π/2 to π/2};
I = (1/2) [ (π)^3/24 + (π)^3/24 ]
I = (π)^3 / 24 = 1.292.

Answer 3

sorry, it's defeated me for the moment!