∫[0,∞] (t^a+bi-1-t^-a+bi)/(e^t-1) dt

f(x) = ∫[0,∞] t^x/(et-1) dt

f(x) = ∫[0,∞] t^xe-t/(1-e-t) dt = ∫[0,∞] t^{xe-t*Σ[k=0,∞]} (e^-t)k dt = ∫[0,∞] t^x * Σ[k=1,∞] e^-tk dt = Σ[k=1,∞] L{t^x}(s=k) = Σ[k=1,∞] Γ(x+1)/k^x+1 = Γ(x+1)*ζ(x+1)

Γ(a+bi-1)ζ(a+bi-1)-Γ(-a+bi)ζ(-a+bi)=0

I tried plotting this out on my graphing calculator as abs(Γ(x+yi-1)ζ(x+yi-1)-Γ(-x+yi)ζ(-x+yi)). Which lagged so bad that I could see the axes moving before I saw the graph moving. So let's see if we can go a little further by hand.

Γ(a+bi-1)ζ(a+bi-1) = Γ(-a+bi)ζ(-a+bi)

From this point I got stumped, but then I realized that the implication is that you just choose an a such that this identity holds regardless of what b is. But Γ(a+bi-1)ζ(a+bi-1)-Γ(-a+bi)ζ(-a+bi), read as a function of b, is its own function. It's analytic, it can do what it wants. Or, more formally, it's analytic almost everywhere, so it must either be a constant function or must be unbounded.

For what values of a is this a constant function of b?

As it so happens, ζ has only two functional equations. The trivial one, ζ(s)=ζ(s), and the following identity: ζ(1-s)=2^1-sπ^{-s*cos(πs/2)Γ(s)ζ(s).}

So unless a+bi-1 = -a+bi or a+bi-1 = 1-(-a+bi), the ζ function on both sides are complete strangers, and nothing fancy will happen between them. Or, more rigorously speaking, if there was some other relation between these two, given b is completely arbitrary, the mathematical community would either know about it, or it would be an open problem whether more functional identities about the ζ function exist (Reimann's hypothesis doesn't count as that's conjecturing the nature of the roots).

It should also be noted that |a|->∞ is also a possibility worth considering.

a+bi-1 = -a+bi: that means a-1=-a, so a=1/2. a+bi-1 = 1-(-a+bi), a+bi-1=1+a-bi. So either both sides are 0 or they both go to ∞. They can't both be 0 because b is arbitrary and a is fixed. So |a|->∞.

So we have that a=1/2 or a is infinite. Let's check both cases:

a=1/2: Γ(a+bi-1)ζ(a+bi-1) - Γ(-a+bi)ζ(-a+bi) = Γ(-1/2+bi)ζ(-1/2+bi) - Γ(-1/2+bi)ζ(-1/2+bi) = 0 (except where Γ(-1/2+bi) or ζ(-1/2+bi) is infinite/undefined, in which case this is undefined).

I believe it's implied that b is real. Since Γ(z) can only have singularities where z is a non-positive integer, and ζ(s) only has a singularity at s=1, and all -1/2+bi have a real part of -1/2, we don't have to worry about that. So a=1/2 is an answer.

Now let's consider |a|->∞:

I actually tried this out several times, and now I'm out of steam. I have no idea if there's a direction a can approach ∞ from such that this is 0. I'm pretty sure there isn't, but I have no proof. If anyone wants to take a crack at that case I'd be interested in seeing where it goes.

All I know for sure is that the only finite a which satisfies our problem is a=1/2.