Integral entailing a complex component Let $a,\sigma,t >0$ and $\mu,\omega \in \mathbb{R}$; $i$ is the imaginary symbol. Looking for the integral;
$$\chi(\omega)=\frac{1}{\sqrt{2 \pi } a \sigma  \sqrt{t} }\int_0^1 -\frac{\exp \left(\mathbf{i} x \omega-\frac{\left(-a \mu +\log \left(\frac{1}{1-x}\right)+\log (x)\right)^2}{2 a^2 \sigma ^2} \right)}{(x-1) x}\, dx$$
I tried integration by parts and substitution to no avail.
 A: Trace of solution, don't really know if that may help but we try...
First of all let's rewrite the integral forgetting about the term in front of it and the minus sign, that is I will ignore the term
$$-\frac{1}{\sqrt{2\pi t}\sigma a}$$
Now, let's rewrite the integral with the easier writing:
$$A = a\mu ~~~~~~~ B = 2a^2\sigma^2$$
And let's manipulate the logarithms as follow:
$$\ln\left(\frac{1}{1-x}\right) = -\ln(1-x)$$
so
$$-\ln(1-x) + \ln(x) = \ln\frac{x}{1-x}$$
Trivial log property.
The integral now is
$$\large \int_0^1 \large \frac{1}{x(1-x)} \large e^{i x\omega}\ e^{\large \left(\frac{\large \left(-A + \ln\frac{x}{1-x}\right)^2}{B}\right)}\ \text{d}x$$
Now let's prefer the change of variable
$$z = -A + \ln\frac{x}{1-x}$$
With an easy calculation you will find that
$$\text{d}z = \frac{\text{d}x}{x(1-x)}$$
And the inverse transformation yelds to
$$x = \frac{e^{z+A}}{1 + e^{z+A}}$$
The new integral now runs like
$$\large \int_{-\infty}^{+\infty} e^{-i\omega \large \frac{e^{z+A}}{1 + e^{z+A}}} e^{\frac{z^2}{B}}\ \text{d}z$$
Then I get stuck, ish, too.
I have some ideas but I first need to check their validity and I'm on the train now, so it's not very comfortable...
CRAZY IDEA
Given the last integrale, I thought about this: There could exist two values $M_1$ and $M_2$ such that e can split
$$\int_{-\infty}^{+\infty} = \int_{-\infty}^{M_1} + \int_{M_1}^{M_2} + \int_{M_2}^{+\infty}$$
And those values are such that both the integrals
$$\int_{-\infty}^{M_1}$$
and
$$\int_{M_2}^{+\infty}$$
lead to suppose that within those ranges, we can approximate a term in the integral:
$$\frac{e^{z+A}}{1 + e^{z+A}} \approx 1$$
hence we would get
$$\int_{-\infty}^{M_1} e^{i\omega} e^{\frac{z^2}{B}}\ \text{d}z$$
Which is solvable in terms of the imaginary error function:
$$e^{i\omega}\frac{1}{2} \sqrt{\pi } \left(\sqrt{B} \text{erfi}\left(\frac{M_1}{\sqrt{B}}\right)+\sqrt{-B}\right)$$
And in the same way
$$\int_{M_2}^{+\infty} e^{i\omega}e^{\frac{z^2}{B}}\ \text{d}z$$
Which is
$$\frac{1}{2} \sqrt{\pi } \left(\sqrt{-B}-\sqrt{B} \text{erfi}\left(\frac{M_2}{\sqrt{B}}\right)\right)$$
So we would get the partial result
$$\frac{\sqrt{\pi}e^{i\omega}}{2}\left\{2\sqrt{-B} + \sqrt{B}\left(\text{erfi}\left(\frac{M_1}{\sqrt{B}}\right) - \text{erfi}\left(\frac{M_2}{\sqrt{B}}\right)\right)\right\}$$
With the remaining questions:

*

*What are $M_1$ and $M_2$?


*what is also ?
$$\int_{M_1}^{M_2} e^{-i\omega \large \frac{e^{z+A}}{1 + e^{z+A}}} e^{\frac{z^2}{B}}\ \text{d}z$$
I smell something like Laplace method.. or the stationary phase.. but I need more time.
I'm sorry if that seems like an insane idea lol, but it's what I thought
