I'm solving the wave equation $$ \begin{cases} \dfrac{\partial^2 u}{\partial t^2} - v^2 \dfrac{\partial^2 u}{\partial x^2} =0, &(x,t) \in (0;1) \times \mathbb{R}^+_{*},\\ u(t,0) =u(t,1) =0, & t \in \mathbb{R}^+_{*},\\ u(0,x) = e^{-400(x-0.3)^2}, &x\in (0;1),\\ \dfrac{\partial u}{\partial t}(0,x) = 0, & x\in (0;1). \end{cases} $$
By using the Fouriers series to solve, the solution takes form (corrected): \begin{align*} u(t,x) = \sum_{h=1}^{\infty}a_h\cos(hv\pi t)\sin(h\pi x), \end{align*} with \begin{align*} a_h = 2\int_{0}^{1}e^{-400(x-0.3)^2}\sin(h\pi x) dx. \end{align*} But I'm having trouble calculating this integral, $$a_h = 2\int_{0}^{1}e^{-400(x-0.3)^2}\sin(h\pi x) dx. $$ with $h \in \mathbb{N}, h \ge 1$. Could someone please help me calculate this integral?
With $$f(x)\mapsto\exp \left(-k^2 (x-\xi )^2\right)\sin (\pi h x)$$
we get the antiderivative
$$F(x)\mapsto\int f(x) \, dx=-\frac{\sqrt{\pi } e^{-\frac{1}{4} \pi h \left(\frac{\pi h}{k^2}+4 i \xi \right)} \left(e^{2 i \pi h \xi } \text{erfi}\left(\frac{\pi h}{2 k}+i k (x-\xi )\right)+\text{erfi}\left(\frac{\pi h}{2 k}-i k (x-\xi )\right)\right)}{4 k}$$
Now we compute the definite integral $$I_1=\int_0^1 f(x) \, dx=F(1)-F(0)=\frac{\sqrt{\pi } e^{-\frac{1}{4} \pi h \left(\frac{\pi h}{k^2}+4 i \xi \right)} \left(-\text{erfi}\left(\frac{\pi h}{2 k}+i k (\xi -1)\right)-e^{2 i \pi h \xi } \text{erfi}\left(\frac{\pi h}{2 k}-i k (\xi -1)\right)+\text{erfi}\left(\frac{\pi h}{2 k}+i k \xi \right)+e^{2 i \pi h \xi } \text{erfi}\left(\frac{\pi h}{2 k}-i k \xi \right)\right)}{4 k}$$
The final result then is $$a_h=2 I_1 |_ {k=20,\xi=\frac{3}{10}}=\frac{1}{40} i \sqrt{\pi } e^{-\frac{\pi h (\pi h+480 i)}{1600}} \left(-e^{\frac{3 i \pi h}{5}} \text{erf}\left(6+\frac{i \pi h}{40}\right)+\text{erf}\left(14+\frac{i \pi h}{40}\right)+\text{erf}\left(6-\frac{i \pi h}{40}\right)-e^{\frac{3 i \pi h}{5}} \text{erf}\left(14-\frac{i \pi h}{40}\right)\right)$$
with erf(z) the error function.
Animation of $u(t,x)$: