integration by parts with normal and complex Hey could someone help me to compute the following integrals, I have two results in front of me, so maybe it would be helpful to see your steps! thanks a lot:
We know that the Euler scheme of a forward SDE is given as follows, 
\begin{align*}
    X_{i+1}^\pi= X_{i}^\pi + b \Delta t + \sigma \Delta W_i 
\end{align*}
Suppose we want to approximate the expectation
we need to approximate the expectations of the form:
 \begin{equation}
    \label{FCB-1}
    I_0:= \mathbb{E}\left [v(t_{i+1},X_{i+1}^{\pi})\Delta W_i|X_{i}^{\pi}=x \right]
\end{equation}
where $v$ represents a general function.\
In the next step we use the Fourier-Cosine method on $v(t_{i+1},\xi)$, we get
\begin{align*}
    v(t_{i+1},\xi)&= \sum_{k=0}^{\infty} \text{'} V_k \cos \left(k \pi \frac{\xi -a}{b-a} \right)\\
    &\approx \sum_{k=0}^{N} \text{'} V_k \cos \left(k \pi \frac{\xi -a}{b-a} \right)
\end{align*}
This gives us the following approximation, 
\begin{align*}
   & I_1= \sum_{k=0}^{N} \text{'} V_k \mathbb{E} \left[ \cos \left(k \pi \frac{\xi -a}{b-a}\right) \Delta W_i\bigg \lvert X_i^{\pi }=x \right] \\
   \text{with }& V_k(t_{i+1})=\frac{x}{b-a} \int_a^b v(t_{i+1},\xi)\cos \left(k \pi \frac{\xi -a}{b-a} \right) d \xi.
\end{align*}
From $I_1$ we will rewrite the expectation, 
\begin{align*}
   \mathbb{E} &\left[ \cos \left(k \Pi \frac{\xi -a}{b-a}\right) \Delta W_i \bigg \lvert X_i^{\pi }=x \right]\\
   &= \mathcal{R}\left \{ \mathbb{E}\left[exp \left( ik \pi \frac{x+ b\Delta t+ \sigma \Delta W_i-a}{b-a}  \right)   \Delta W_i \bigg \lvert X_i^{\pi }=x \right]  \right\}\\
    &= \mathcal{R}\left \{ \mathbb{E}\left[exp \left( ik \pi \frac{x+ b\Delta t-a}{b-a}  \right) \Delta W_i exp \left( ik \pi \frac{ \sigma \Delta W_i}{b-a}  \right)    \right]   \right\}\\
      &= \mathcal{R}\left \{ exp \left( ik \pi \frac{x+ b\Delta t-a}{b-a}  \right)\mathbb{E} \left[ \Delta W_i exp \left( ik \pi \frac{ \sigma \Delta W_i}{b-a}  \right) \right]   \right\}
      \end{align*}
We focus on the last term 
\begin{align*}
\mathbb{E} \left[ \Delta W_i exp \left( ik \pi \frac{ \sigma \Delta W_i}{b-a}  \right) \right] =
\frac1{\sqrt{2 \pi \Delta_t}} \int_{- \infty}^{\infty} y\, \text{exp}\left(ik\pi \frac{\sigma y}{b-a}\right)\, \text{exp}\left(- \frac{y^2}{\Delta_ t}\right)\,\mathrm d y
\end{align*}
and from there i am stuck :S, the $\Delta W_i \sim N(0,\Delta t) $
The final result/approximation of I_0 should be $\sum_{k=0}^{N-1}V_k(t_{i+1})\sigma \Delta t \mathcal{R}\left( i \frac{k \pi}{b-a}\phi(\frac{k \pi}{b-a})exp {ik\pi \frac{x-a}{b-a
}}\right)$
 A: Starting with
$$I = \frac1{\sqrt{2 \pi \Delta_t}} \int_{- \infty}^{\infty} y\, \text{exp}\left(ik\pi \frac{\sigma y}{b-a}\right)\, \text{exp}\left(- \frac{y^2}{\Delta_ t}\right)\,\mathrm d y$$
make the following substitutions
$$y = \sqrt{\pi \Delta_t} x$$
$$s = k \dfrac{\sqrt{\pi \Delta_t}}{2}\dfrac{\sigma}{a-b}$$
and rewrite the integral such that it is easily recognizable as a Fourier Transform
$$I = \dfrac{\sqrt{\pi \Delta_t}}{\sqrt{2}}\int_{- \infty}^{\infty} x\, \text{exp}\left(-\pi x^2\right)\, \text{exp}\left(-2\pi i xs\right)\,\mathrm d x$$
$$ = -\dfrac{1}{2\pi i} \dfrac{\sqrt{\pi \Delta_t}}{\sqrt{2}}\int_{- \infty}^{\infty} -2\pi ix\, \text{exp}\left(-\pi x^2\right)\, \text{exp}\left(-2\pi i xs\right)\,\mathrm d x$$
$$ = -\dfrac{1}{2\pi i} \dfrac{\sqrt{\pi \Delta_t}}{\sqrt{2}}F\left\{-2\pi ix\, \text{exp}\left(-\pi x^2\right)\right\}$$
so by the derivative property of the Fourier Transform
$$ I = -\dfrac{1}{2\pi i} \dfrac{\sqrt{\pi \Delta_t}}{\sqrt{2}} \dfrac{\mathrm d}{\mathrm{d}s}F\left\{\text{exp}\left(-\pi x^2\right)\right\}$$
and looking up the Fourier Transform of the Gaussian in a table of Fourier Transforms
$$ I = -\dfrac{1}{2\pi i} \dfrac{\sqrt{\pi \Delta_t}}{\sqrt{2}} \dfrac{\mathrm d}{\mathrm{d}s}\left[\text{exp}\left(-\pi s^2\right)\right]$$
$$ = -i \dfrac{\sqrt{\pi \Delta_t}}{\sqrt{2}} s \, \text{exp}\left(-\pi s^2\right)$$
and then finally substituting back in for $s$ and performing a little algebra 
$$ I = i \dfrac{\pi \Delta_t}{2\sqrt{2}}\dfrac{\sigma}{b-a} k \, \text{exp}\left(-\dfrac{\pi^2}{4}\left[\dfrac{\sigma}{b-a}\right]^2 \Delta_t k^2\right)$$
You should double check the algebra on that last step for mistakes. 
