show that $\lim_{r\to\infty} \int_0^2 g(x) \exp^{ikrx} dx = 0$ let g(x) be a sufficiently smooth function and k,r $\in \rm I\!R$ 
via partial integration ($\int uv' = [uv] - \int u'v$), I arrive at
$$I:\lim_{r\to\infty}\left( \left(\frac{g(2)}{ikr}\exp^{2ikr}-\frac{g(0)}{ikr}\right)-\left[  \int_0^2 \frac{1}{(ikr)}g'(x) \exp^{ikrx} dx\right]\right)$$
When I then choose g'(x) to be v' and the exponential term to be u, I get
$$II:\lim_{r\to\infty}\left( \left(\frac{g(2)}{irk}\exp^{2ikr}-\frac{g(0)}{ikr}\right)-\left[ \left[g(x) \frac{1}{ikr} \exp^{ikrx} \right]^2_0 -\int_0^2 g(x) \exp^{ikrx} dx\right]\right)$$
The terms in front of the integral cancel each other and I arrive back at the start. When I choose u & v' the other way arround I'll get
$$III:\lim_{r\to\infty}\left( \left[g(x) \frac{1}{ikr} \exp^{ikrx} \right]^2_0-\left[ \left[g'(x) \frac{1}{(ikr)^2} \exp^{ikrx} \right]^2_0 -\int_0^2 g''(x) \frac{1}{(ikr)^2} \exp^{ikrx} dx\right]\right)$$
I don't see how this approach could bring me further, since the integral didn't simplify. I haven't used the limit $\lim_{r\to\infty}$ yet, so I guess this should be rather important. But on the other hand side, I also don't see how the limit would help, since the exponential function will always go to infinity faster than the fraction.
I missed that the exponential function is oscilating, so it will go to zero for r going to infinity. This clears removes the terms in front of the integral in III. For the integral $\int_0^2 g''(x) \frac{1}{(ikr)^2} \exp^{ikrx} dx$, can I argue that it will also go to zero, since the integral of $exp^{ikr}$ will always be oscilating an the prefactor $\frac{1}{(ikr)}$ overpowers the rest?
 A: Let $f_{g,k}(r):=\int_0^2g(x)e^{ikrx}dx$. Then if I understand the question right, we want to show that $f_{g,k}(r)\to_{r\to\infty} 0$ for all $k\in\Bbb R\setminus \{0\}$ (not true for $k=0$!) and for all sufficiently smooth $g:[0,2]\to \Bbb C$.

First, let $g:[0,2]\to\Bbb R$ and $k>0$.
Note that
$$f_{g,k}(r) = \int_0^2g(x)\cos(krx)dx+i\int_0^2g(x)\sin(krx)dx\text.$$
Since a complex series converges if and only if its real and imaginary parts converge, we need to show that $\Re f_{g,k}(r)\to_{r\to\infty} 0$ and $\Im f_{g,k}(r)\to_{r\to\infty} 0$.
We show $\Im f_{g,k}(r)\to_{r\to\infty} 0$. The same argument works for $\Re f_{g,k}$.
Let $r>\frac{\pi}k$. Then we split the integral to parts according to the periodicity of the $\sin$ function. The idea being that the sum is "almost $0$" because each part is. Formally:
$$\Im f_{g,k}(r) = \int_0^2g(x)\sin(krx)dx=\\
\int_0^\frac{2\pi}{kr}g(x)\sin(krx)dx+\int_\frac{2\pi}{kr}^\frac{4\pi}{kr}g(x)\sin(krx)dx+\cdots+\int_\frac{2N\pi}{kr}^2g(x)\sin(krx)dx=\\
\sum_{n=1}^N\underbrace{\int_\frac{(2n-2)\pi}{kr}^\frac{2n\pi}{kr}g(x)\sin(krx)dx}_{(*)}+\underbrace{\int_\frac{2N\pi}{kr}^2g(x)\sin(krx)dx}_{(**)}$$
where $N\in\Bbb N$ is naturally selected such that $\frac{2N\pi}{kr}\le 2<\frac{(2N+2)\pi}{kr}$ (namely, $N=\left\lfloor\frac{kr}\pi\right\rfloor$).
Then,
$$|(**)|=\left|\int_\frac{2N\pi}{kr}^2g(x)\sin(krx)dx\right|\le\int_\frac{2N\pi}{kr}^2\left|g(x)\sin(krx)\right|dx\le\int_\frac{2N\pi}{kr}^2\left|g(x)\right|dx\le\\
\left(2-\frac{2N\pi}{kr}\right)\max|g|<\left(\frac{(2N+2)\pi}{kr}-\frac{2N\pi}{kr}\right)\max|g|=2\max|g|\frac\pi{kr}\text.$$
Since $g'$ is continuous on $[0,2]$, there exists $M:=\max g'([0,2])$. So,
$$|(*)|=\left|\int_\frac{(2n-2)\pi}{kr}^\frac{(2n-1)\pi}{kr}g(x)\sin(krx)dx+
\int_\frac{(2n-1)\pi}{kr}^\frac{2n\pi}{kr}g(x)\sin(krx)dx\right| \underset{x=y+\pi/kr}{=}\\
\left|\int_\frac{(2n-2)\pi}{kr}^\frac{(2n-1)\pi}{kr}g(x)\sin(krx)dx+\int_\frac{(2n-2)\pi}{kr}^\frac{(2n-1)\pi}{kr}g\left(y+\frac\pi{kr}\right)\underbrace{\sin(kry+\pi)}_{=-\sin(kry)}dy\right|=\\
\left|\int_\frac{(2n-2)\pi}{kr}^\frac{(2n-1)\pi}{kr}\sin(krx)\cdot\left(g(x)-g\left(x+\frac\pi{kr}\right)\right)dx\right|\le\int_\frac{(2n-2)\pi}{kr}^\frac{(2n-1)\pi}{kr}\sin(krx)\left|g(x)-g\left(x+\frac\pi{kr}\right)\right|dx\le\\
\int_\frac{(2n-2)\pi}{kr}^\frac{(2n-1)\pi}{kr}\sin(krx)\cdot|M|\cdot\left|x-\left(x+\frac\pi{kr}\right)\right|dx=2|M|\frac\pi{k^2r^2}=\frac{2N\pi}{kr}\cdot\frac{|M|}{Nkr}\le2\frac{|M|}{Nkr}\text,$$ which yields
$$\left|\sum_{n=1}^N(*)\right|\le\sum_{n=1}^N|(*)|\le\sum_{n=1}^N2\frac{|M|}{Nkr}=2\frac{|M|}{kr}\text.$$
Putting it all together,
$$\left|\Im f_{g,k}(r)\right|=\left|\sum_{n=1}^N(*)+(**)\right|\le\left|\sum_{n=1}^N(*)\right|+\left|(**)\right|<2\frac{|M|}{kr}+2\max|g|\frac\pi{kr}\to_{r\to\infty}0\text.$$

Now, let $k<0$. Then $$f_{g,k}(r) = \int_0^2g(x)\cos(krx)dx+i\int_0^2g(x)\sin(krx)dx=\\
\int_0^2g(x)\cos(-krx)dx-i\int_0^2g(x)\sin(-krx)dx\to_{r\to\infty} 0$$
as shown.

Finally, let $g:[0,2]\to\Bbb C$. Then $$f_{g,k}(r)=\int_0^2g(x)e^{ikrx}dx=\int_0^2\Re g(x)e^{ikrx}dx+i\int_0^2\Im g(x)e^{ikrx}dx\to_{r\to\infty}0+i\cdot 0=0$$
applying the above result to $\Re g$ and $\Im g$, which are real-valued.
