# Thread: counter integral

1. ## counter integral

Let y(x) be a real valued function defined on the interval $0 by means of the equations:

$y(x) = \left\{ \begin{array}{rcl}
x^3 \sin(\pi/x) & \mbox{for} & 0 0 & \mbox{for} & x=0
\end{array}\right.
$

show that the equation $z=x+iy(x) \ (0\leq x \leq 1)$ representing an arc $C$ that intersects the real axis at the point $z=\frac{1}{n} \ (n=1,\ 2\, \ \dotso)$ and $z=0$

I'm not sure if I'm picking the right values for this, but I would think that it would something along the lines of:

$\int_a^b f[z(t)]z^,(t) dt$

$=\int_0^1 \bigg{(} x+i(x^3 \sin(\pi/x)) \bigg{)}\cdot\left(\frac{d}{dx} x^3 \sin(\pi/x)) \right) dx$ this is the part that I'm not sure of, but continuing would give me:

$\int_0^1 \bigg{(} x+i(x^3 \sin(\pi/x)) \bigg{)} \cdot \bigg{(}3x^2\sin(\pi/x)-x\cos(\pi/x)\cdot\pi \bigg{)} dx$

$\int_0^1 3x^3\sin(\pi/x)-x^2\pi\cos(\pi/x)+i(3x^5\sin^2(\pi/x)-x^4\pi\sin(pi/x)\cos(\pi/x)) \ dx$

putting the above equation into Maple gave me:

$-\frac{1}{48} \pi^5 + \frac{1}{24} \cdot Si(\pi) \pi^4 - \frac{1}{24} \pi^3+\frac{1}{12}\pi$ for some reason this doesn't look correct.

2. Your method of attack seems very interesting but is unnecessary.
Note that for x=0
$z=0$
which is one of the intersections.
Now for x>0 we have
$z=x+x^3sin(\frac{\pi}{x})i$.
This intersects the real axis when the imaginary component is zero. This happens when
$x^3sin(\frac{\pi}{x})=0$.

Now solve for x.