What is the value of $\frac{1}{2\pi}\int_{-\infty}^{\infty}\prod_{k=1}^\infty\Big(1-p+p\exp \frac{it}{2^k}\Big)dt$? Here $0<p<1$. If $p=\frac{1}{2}$ the value is $1$. In all cases the value is a positive real number (or $+\infty$). This integral is associated with the inverse Fourier transform of some characteristic function. For details, see here and here. 
Clarification:
Consider $\phi(t)$ to be the characteristic function of an infinite weighted sum of i.i.d. Bernouilli($p$) random variables, the weights being $\frac{1}{2^k}, k= 1, 2, \cdots$. This Fourier transform is an infinite product by virtue of the convolution theorem. Now consider $f(x)$, the derivative of the inverse Fourier transform of $\phi(t)$: $f(x)$ is a density on $[0, 1]$ and if $p=0.5$ it is the uniform density. The integral in my question is $f(0)$. I am wondering if and how it varies depending on $p$ (and of course, if $p=1/2$, then $f(0) = 1 = f(x)$ for all $x\in [0, 1]$).
 A: Your product comes from the following sequence of convolution operators
Let $p \in \Bbb{C}$ and $$T_0f = f, \qquad T_kf = T_{k-1} f+p(T_{k-1} f(.+\frac1{2^k})-T_{k-1} f)$$
The first thing to check is the convergence of $\lim_{k \to \infty} T_k f$.
If $f$ is a polynomial of degree $d$ then so is $T_k f$, let $$\| \sum_{n=0}^d c_n x^n\| = \sum_{n=0}^d |c_n|$$
then $$ \| p ((x+2^{-k})^n-x^n)\|=\|p\sum_{m=1}^n {n\choose m} x^{n-m}2^{-km}\| = |p| \sum_{m=1}^n {n \choose m} 2^{-km}= O(2^{-k})$$
Thus $\sum_{k \ge 1} \|T_k f-T_{k-1}f\| = \sum_{k \ge 1} O(2^{-k})<\infty$ so that $(T_k f)_{k \ge 0}$ is a Cauchy sequence and it converges to $T_\infty f=\lim_{k \to \infty}T_k f $ a polynomial of degree $d$, the convergence being locally uniform.
$T_k f = f \ast h_k$ where $h_k$ is a sum of shifted Dirac deltas, I don't know why you'd except that $h_k(0)$ and $h_\infty(0)$ would be defined.
It is not obvious at all if $\lim_{k \to \infty}T_k f $ converges for any non-polynomial analytic function. 
The Fourier transform of $h_\infty$ is $H_\infty(t)=\prod_{k \ge 1} (1+p (e^{it 2^{-k}}-1))$ which is smooth but we also need to study its growth to deduce something on $h_\infty$. It is quite obvious $\int_{-\infty}^\infty H_\infty(t)dt$ diverges.
A: I will show that the only meaningful value of the above product occurs at $p=1/2$.  First, at $p=1/2$, the product is equal to
$$\begin{align} \prod_{k=1}^{\infty} \left ( \frac12 + \frac12 e^{i t/2^k} \right ) &= \prod_{k=1}^{\infty} \left ( \frac12 e^{-i t/2^{k+1}} + \frac12 e^{i t/2^{k+1}} \right ) e^{i t/2^{k+1}} \\ &= \prod_{k=1}^{\infty}  \cos{\left (\frac{t}{2^{k+1}} \right )} \prod_{k=1}^{\infty} e^{i t/2^{k+1}} \\ &= \frac{\sin{(t/2)}}{t/2} e^{i t/2}\end{align}  $$
Accordingly, the integral is
$$\begin{align} \int_{-\infty}^{\infty} dt \, \frac{\sin{(t/2)}}{t/2} e^{i t/2} &= 2 \int_{-\infty}^{\infty} dt \, \frac{\sin{(t/2)} \cos{(t/2)}}{t} \\ &= \int_{-\infty}^{\infty} dt \, \frac{\sin{t}}{t} \\ &= \pi \end{align} $$
I will now show that, for a small perturbation away from $p=1/2$, the integral of the perturbation term does not converge to a finite value.  To see this, we evaluate the product for a small perturbation $\epsilon$ from $p=1/2$:
$$\begin{align} \prod_{k=1}^{\infty} \left [ \left ( \frac12 - \epsilon \right ) + \left (\frac12 + \epsilon \right ) e^{i t/2^k} \right ] &=  \prod_{k=1}^{\infty} \left [ \left ( \frac12 + \frac12 e^{i t/2^k} \right ) - \epsilon \left ( \frac12 - \frac12 e^{i t/2^k} \right )  \right ] \\ &= \prod_{k=1}^{\infty} \left ( \frac12 + \frac12 e^{i t/2^k} \right ) - \epsilon \sum_{m=1}^{\infty} \left ( \frac12 - \frac12 e^{i t/2^m} \right ) \times \\ & \prod_{k=1\\k \ne m}^{\infty} \left [ \left ( \frac12 + \frac12 e^{i t/2^k} \right ) \right ] + O \left ( \epsilon^2 \right ) \\ &= \frac{\sin{(t/2)}}{t/2} e^{i t/2} + i \epsilon \frac{\sin{(t/2)}}{t/2} e^{i t/2} \sum_{m=1}^{\infty} \tan{\left ( \frac{t}{2^{m+1}} \right )}  \\ & + O \left ( \epsilon^2 \right ) \end{align}$$
We consider the perturbation term, which is the original product times the sum of the tangent functions.  This sum converges over values of $t$ not equal to a power of $2$ times $\pi$.  Nevertheless, the integral over $t$ involves a $\sin{t}$ times the tangents, and the product of that sine and tangent is finite for all $t$.  The integral of the perturbation term is then
$$i \int_{-\infty}^{\infty} dt \, \frac{\sin{(t/2)}}{t/2} e^{i t/2} \sum_{m=1}^{\infty} \tan{\left ( \frac{t}{2^{m+1}} \right )} $$
The sum converges absolutely, and the integral of each term in the sum is finite as we will see.  Therefore, by the monotone convergence theorem, we may reverse the order of the integral and sum (actually, the order of the integral and limit as the number of terms in the sum becomes infinite).  Accordingly, the integral is equal to the following sum:
$$-\sum_{m=1}^{\infty} \int_{-\infty}^{\infty} dt \, \frac{\sin^2{(t/2)}}{t/2} \tan{\left ( \frac{t}{2^{m+1}} \right )} = -2 \sum_{m=1}^{\infty} \int_{-\infty}^{\infty} dt \, \frac{\sin^2{t}}{t} \tan{\left ( \frac{t}{2^{m}} \right )}$$
We now evaluate each term in the sum, i.e., each integral over $t$.  I find it easier to rewrite the integrals using
$$\sin{t} \tan{\left ( \frac{t}{2^{m}} \right )} = 2^m \cos{ \frac{t}{2}} \cos{ \frac{t}{4}} ... \cos{ \frac{t}{2^{m-1}}} \sin^2{\frac{t}{2^m}} $$
Each integral is then equal to
$$-2^m \int_{-\infty}^{\infty} dt \, \frac{\sin{t}}{t} \cos{ \frac{t}{2}} \cos{ \frac{t}{4}} ... \cos{ \frac{t}{2^{m-1}}} \sin^2{\frac{t}{2^m}} $$
I will show that the integral is independent of $m$.  To simplify the analysis to follow, I rewrite the integral as
$$-2^m \int_{-\infty}^{\infty} dt \, \frac{\sin{2^m t}}{t} \cos{ 2^{m-1} t} \, \cos{ 2^{m-2} t} ... \cos{ 2 t}\, \sin^2{t} $$
This will seem like lunacy, but I will now expand each trig function into complex exponentials.  We then integrate the result in the complex plane, replacing real $t$ with complex $z$; the positive exponentials each produce a contribution of $i \pi$ and the negative exponentials each produce a contribution of $-i \pi$.  We then add up the contributions (times the coefficients of each exponential) and divide the result by $-i 2^{m+2}$.
To keep the discussion a bit concrete, I will illustrate an example of $m=3$.  The expansion into exponentials looks like
$$\left (e^{i 8 z} - e^{-i 8 z} \right ) \left (e^{i 4 z} + e^{-i 4 z} \right ) \left (e^{i 2 z} + e^{-i 2 z} \right ) \left (e^{i 2 z} + e^{-i 2 z} - 2 \right ) $$
Expanding this out, the result is (where $c.c.$ stands for complex conjugate):
$$\left ( e^{i 16 z} + 2 e^{i 12 z} + 2 e^{i 8 z} + 2 e^{i 4 z} - c.c. \right ) - 2 \left (e^{i 14 z} + e^{i 10 z} + e^{i 6 z} + e^{i 2 z} - c.c. \right )$$
The imbalance of terms results from a cancellation of a constant term and its complex conjugate.  The contributions from each of the exponentials to the integral are then $i 14 \pi - i 16 \pi = -i 2 \pi$.  Accordingly, the integral for $m=3$ is $\pi/2$.
Note that the imbalance in exponentials is the case for all $m$.  The first set of parentheses will always have one pair of exponentials less than the second set of parentheses because of the single cancellation of the constant term of its complex conjugate.  Therefore, we can say that
$$\int_{-\infty}^{\infty} dt \, \frac{\sin^2{t}}{t} \tan{\left ( \frac{t}{2^{m}} \right )} = \frac{\pi}{2} $$
independent of the value of $m$, when $m \in \mathbb{N}$.  Because each integral in the summand is equal to $\pi/2$, the infinite sum of the integrals is accordingly an infinite number of $\pi/2$'s.  Therefore, the integral over the perturbation term is infinite, and the only meaningful value of the integral is for $p=1/2$.
