Is my expression too complicated for FullSimplify or am I doing something wrong?

Question

I have a messily defined function $v(h, w)$ with $h, w \in \mathbb{R}$ and with a removable singularity at $h=1/2$, and I am trying to prove some of its properties using Mathematica. In particular I want to prove that $v(h,w) \in \mathbb{R}$, as opposed to possibly having nonzero imaginary part, and I want to prove that $v(h,w) = v(1-h, -w)$. I am defining the function in Mathematica as follows:

a[h_] = h / (2*h - 1)
q[h_, w_] = Sqrt[2*h - 1] * Sqrt[2*w]
v[h_, w_] = Exp[-w]*(2/Sqrt[Pi])*q[h,w]*Exp[(q[h,w]*a[h])^2] / ( Erfi[q[h,w]*a[h]] + Erfi[q[h,w]*(1 - a[h])] )

I apologize for its messiness. My initial hope was that Mathematica would tell me that this is just a well known special function with known properties which I could cite. My next hope was that I could at least use applications of FullSimplify with Assumptions to prove the properties. But I have not been successful at either approach.

To make my question more concrete, I am wondering why Mathematica does not simplify the following expression to True:

FullSimplify[ v[h, w] == v[1 - h, -w], Element[{w,h}, Reals] && h != 1/2]

One possibility is that I am wrong about the math. I've tried to rule this out by using Mathematica to prove special cases like the following, where I've also tried other values of $h$ besides $1/5$.

FullSimplify[ v[1/5, w] / v[1 - 1/5, -w] ]
FullSimplify[ v[1/5, w] - v[1 - 1/5, -w] ]
FullSimplify[ v[1/5, w] == v[1 - 1/5, -w] ]

Another possibility is that it is too hard for Mathematica. But I think the most likely explanation is that I'm using Mathematica wrong because I'm a newbie.

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Update:
Inspired by @rcollyer's answer I took out a pencil and paper and figured out how to write this function as the reciprocal of a series that satisfies the properties by construction:

fterm[h_, w_, k_] = (((-4*w)^k) / (2*k+1)!!) * (h / (2*h-1)) * Exp[w]*(h^2 / (2*h-1))^k
f[h_, w_] = Sum[fterm[h, w, k] + fterm[1-h, -w, k], {k, 0, Infinity}]

Mathematica is still unable to directly verify that $f(h,w) = f(1-h,-w)$ on the appropriate domain, but $1/f(h,w)$ compares True to the original function $v(h,w)$. As a side note, I had hoped that in the process of verifying these properties I would find a formula that gets rid of the computational difficulty associated with the removable singularity at $h=1/2$ for free, but this did not happen.

Answer 1

Simplification in Mathematica is often a black art, and requires great use of your own intuition and knowledge to be effective. That said, I bring your attention to the series from of $\DeclareMathOperator{\erfi}{erfi}\erfi(z)$,

$$\erfi(z) = \frac{2}{\sqrt{\pi}}\sum^\infty_{k=1} \frac{z^{(2k+1)}}{k! (2k+1)}.$$

Consider what happens when we use that to combine the terms in the denominator of v[h, w]:

((Sqrt[2] h Sqrt[w])/Sqrt[-1 + 2 h])^(2 k + 1) + 
 (Sqrt[2] Sqrt[-1 + 2 h] (1 - h/(-1 + 2 h)) Sqrt[w])^(2 k + 1)

ignoring the $k! (2k+1)$, for now. FullSimplify will not tackle that even with the assumptions Element[{w,h}, Reals], but PowerExpand will and then you can use FullSimplify to get

(-(1/2) + h)^(-(1/2) - k) ((-1 + h)^(1 + 2 k) + h^(1 + 2 k)) w^(1/2 + k)

Similarly, for the denominator of v[1 - h, -w] you get

(-I) E^(I k Pi) (1/2 - h)^(-(1/2) - k) (-(1 - h)^(1 + 2 k) + (-h)^(2 k) h) w^(1/2 + k)

Note, the similarities.

I will leave the rest of the simplification to you.

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Edit: it turns out I can't leave well enough alone. As I mentioned in a comment, PowerExpand assumes that $h\text{,}w\in\mathbb{R}$ and $h\text{,}w\ge0$. The latter requirement may cause problems, but PowerExpand accepts the option Assumptions to alter the default. Here's what happens to the first denominator when the four possibilities of positivity are considered:

PowerExpand[
 denom1, 
 Assumptions -> LessEqual[#1, #2] && LessEqual[#3, #4]
] & @@@ {{w, 0, h, 0}, {w, 0, 0, h}, {0, w, h, 0}, {0, w, 0, h}} // 
FullSimplify[#, {h, w} ∈ Reals && k ∈ Integers] &

$$-i (-1)^k \left(\frac{1}{2}-h\right)^{-k-\frac{1}{2}} \left(h^{2 k+1}+(h-1)^{2 k+1}\right) w^{k+\frac{1}{2}}$$ $$\left(h-\frac{1}{2}\right)^{-k-\frac{1}{2}} \left(h^{2 k+1}+(h-1)^{2 k+1}\right) w^{k+\frac{1}{2}}$$ $$-i \left(\frac{1}{2}-h\right)^{-k-\frac{1}{2}} \left((h-1) (i (h-1))^{2 k}+h (i h)^{2 k}\right) w^{k+\frac{1}{2}}$$ $$\left(h-\frac{1}{2}\right)^{-k-\frac{1}{2}} \left(h^{2 k+1}+(h-1)^{2 k+1}\right) w^{k+\frac{1}{2}}$$

The code itself isn't terribly complex, but let me walk you through it.

PowerExpand[
 denom1, 
 Assumptions -> LessEqual[#1, #2] && LessEqual[#3, #4]
] &

is a pure function taking 4 arguments labelled #i. This is then applied to a list containing the 4 possible orderings of the conditions, i.e. {w, 0, h, 0} corresponds to w <= 0 && h <= 0 in the Assumptions, using the second short hand form of Apply. Essentially,

f[#1, #2]& @@@ {{a, b}, {c, d}}

becomes

{f[a,b], f[c,d]}

The entire list is then fed into FullSimplify with the assumption of reality included. I tried including the positivity assumptions, too, but it tended to convolute the result.

Applied to the second denominator you get,

$$-i \left(\frac{1}{2}-h\right)^{-k-\frac{1}{2}} \left(h^{2 k+1}+(h-1)^{2 k+1}\right) w^{k+\frac{1}{2}} \left( \begin{array}{cc} \{ & \begin{array}{cc} -(-1)^k & w<0 \\ (-1)^k & \text{True} \\ \end{array} \\ \end{array} \right)$$ $$-i \left(\frac{1}{2}-h\right)^{-k-\frac{1}{2}} \left(h^{2 k+1}+(h-1)^{2 k+1}\right) w^{k+\frac{1}{2}} \left( \begin{array}{cc} \{ & \begin{array}{cc} -(-1)^k & w<0 \\ (-1)^k & \text{True} \\ \end{array} \\ \end{array} \right)$$ $$-i (-1)^k \left(\frac{1}{2}-h\right)^{-k-\frac{1}{2}} \left(h^{2 k+1}+(h-1)^{2 k+1}\right) w^{k+\frac{1}{2}}$$ $$-i (-1)^k \left(\frac{1}{2}-h\right)^{-k-\frac{1}{2}} \left(h^{2 k+1}+(h-1)^{2 k+1}\right) w^{k+\frac{1}{2}}$$

End Edit

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As an additional note, if you look at

Series[v[h, w]/v[1 - h, -w], {h, 0, 5}, {w, 0, 5}]

you get

$$\left(1+O\left(w^{11/2}\right)\right)+h O\left(w^7\right)+h^2 O\left(w^{11/2}\right)+h^3 O\left(w^{11/2}\right)+h^4 O\left(w^{11/2}\right)+h^5 O\left(w^{11/2}\right)+O\left(h^6\right)$$

which provides a strong indication that they are equal.