Show that the function $u(x,y) =e^{(x^2−y^2)}\cos(2xy)$ is harmonic. Find the harmonic conjugate of u, up to the addition of a constant. My question is how do you integrate this function?
My approach: To show that $u(x,y)$ is harmonic it will need continuous 2nd order partial derivatives and that $\frac{d^2u}{dx^2}+\frac{d^2u}{dy^2}=0$. Therefore:
$\frac{du}{dx}=-2y\sin (2xy)e^{(x^2-y^2)}+2x\cos (2xy)e^{(x^2-y^2)}$ and $\frac{d^2u}{dx^2}=-4ye^{(x^2-y^2)}(x\sin (2xy)+y\cos (2xy))+4e^{(x^2-y^2)}x(-y\sin (2xy)+x\cos (2xy))+2\cos (2xy)e^{(x^2-y^2)}$
$\frac{du}{dy}=e^{(x^2-y^2)}y\cos (2xy)-2xe^{(x^2-y^2)}\sin (2xy)$
$\frac{d^2u}{dy^2}=4ye^{(x^2-y^2)}(y\cos (2xy)+2x\sin (2xy))-4e^{(x^2-y^2)}x^2\cos (2xy)-2e^{(x^2-y^2)}\cos (2xy)$
Hence $\frac{d^2u}{dx^2}+\frac{d^2u}{dy^2}=0$ Now to find $v(x,y)$ so v is harmonic to u if the Cauchy Riemann equations are continuous and hold you can see that they are continuous, thus:
$\frac{du}{dx}=\frac{dv}{dy}=-2y\sin (2xy)e^{(x^2-y^2)}+2x\cos (2xy)e^{(x^2-y^2)}$
To find v we integrate $∫$$-2y\sin (2xy)e^{(x^2-y^2)}+2x\cos (2xy)e^{(x^2-y^2)}dy$ which brings me to my question, how do you integrate this function?
Yes, you could integrate by parts.
What was hinted at in the comments is that if you can identify a function $f(z)$ holomorphic and $u=\Re f$ then $v=\Im f.$
Since $u=\Re \exp z^2$, we have $v=\Im \exp z^2 +\textrm{const.}= \exp \left( x^2-y^2\right) \sin 2xy + \textrm{const.}$