Evaluate $\int_\gamma\left(4e^{(y-4x+3)^2}-y\right)dx + \left(-e^{(y-4x+3)^2}+2x\right)dy.$

Question

Evaluate $$I=\int_\gamma\left(4e^{(y-4x+3)^2}-y\right)dx + \left(-e^{(y-4x+3)^2}+2x\right)dy,$$

where $\gamma$ is the curvepiece along the parabola $y=x^2$ from $(1,1)$ to $(3,9)$.

Setting $x=t$ and $y=t^2$ I get $dx=dt$ and $dy=2tdt$ so

$$I=\int_1^3 \left(4e^{(t^2-4t+3)^2}-t^2\right) + \left(-e^{(t^2-4t+3)^2}+2t\right)2t \ dt = \int_1^3\left((4-2t) e^{(t-1)^2(t-3)^2}+3t^2\right)dt.$$

The sad part is that I can't evaluate the integral by elementary methods. However, using wolframalpha I get the correct answer of $I=26$.

What is the trick to this?

Answer 1

The problem is simple if you are aware of the imaginary error function. $$I=\int\left(4e^{(y-4x+3)^2}-y\right)dx= -\frac{1}{2} \sqrt{\pi } \text{erfi}(y-4 x+3)-x y$$ $$J=\int \left(-e^{(y-4x+3)^2}+2x\right)dy=2 x y-\frac{1}{2} \sqrt{\pi } \text{erfi}(y-4 x+3)$$ $$I+J=x y-\sqrt{\pi } \text{erfi}(y-4 x+3)$$ Now $y=x^2$, use the bounds for $x$ and remember that $x^2-4x+3=(x-1)(x-3)$ which makes the $\text{erfi}(.)$ equal to $0$ at each boundary.

Answer 2

I think your steps are correct. Lets end with the evaluation of the integration. I think for this type of integration we put $z=t-\frac{a+b}{2}$ where $a, b $ are lower and upper limit of the integration. $$I=\int_1^3\left((4-2t) e^{(t-1)^2(t-3)^2}+3t^2\right)dt=-2\int_{-1}^1z e^{(z^2-1)^2}dz+27-1=26,$$ since the integrand is odd function so the value is zero.