I try the following two method on finding the harmonic conjugate of $T(x,y)= e^{-y} \sin x$ :
Method 1 : by a method of the book Complex Variables and Applications by Brown and Churchill, Chapter 9 Section 104, $$v(x,y) = \int_{(0,0)}^{(x,y)} -u_t(s, t)\ ds + u_s(s, t)\ dt = \int_{(0,0)}^{(x,y)} e^{-y} \sin x \ dx + e^{-y} \cos x \ dy = -e^{-y} \cos x – e^{-y} \cos x +C = -2e^{-y} \cos x +C$$ which $C$ can be chosen to be zero.
Method 1 : Since $T(x,y)= e^{-y} \sin x$ and its harmonic conjugate must be the real and imaginary parts of an analytic function, respectively, so it can be $f(z)=-ie^{iz} =e^{-y} \sin x – ie^{-y} \cos x.$
So why there is an extra $2$ coefficient in Method 1? Where did I do wrong?
Added. Probably the Method 1 is wrong. It comes from if $F_x(x,y) = P(x,y), \ F_y(x,y) = Q(x,y)$ holds in here. For example the method gives the correct answer for $u=xy$ but again it give a wrong answer for $u=x^3-3xy^2:$ that is $v=6x^2y-y^3$; but the correct one is $v=3x^2y-y^3$. But it looks impossible such a famous book to make a mistake. (???)
Best Answer
You did not compute the line integral correctly.
For $u(x,y)=e^{-y}\sin x$, we have $u_{,1}(x,y)=\dfrac{\partial u(x,y)}{\partial x}=e^{-y}\cos x$ and $u_{,2}(x,y)=\dfrac{\partial u(x,y)}{\partial y}=-e^{-y}\sin x$. Hence taking the straight-line path $\gamma\colon\tau\in[0,1]\mapsto(x\tau,y\tau)$ joining $(0,0)$ to $(x,y)$, we have
\begin{align*} &\int_{(0,0)}^{(x,y)} (-u_{,2},u_{,1})(s, t)\cdot(\mathrm{d}s,\mathrm{d}t)\\ &=\int_0^1 (-u_{,2},u_{,1})(\gamma(\tau))\cdot\dot\gamma(\tau)\,\mathrm{d}\tau\\ &=\int_0^1 (-u_{,2},u_{,1})(x\tau,y\tau)\cdot\dot(x,y)\,\mathrm{d}\tau\\ &=\int_0^1 (e^{-y\tau}\sin (x\tau),e^{-y\tau}\cos x\tau)\cdot\dot(x,y)\,\mathrm{d}\tau\\ &=\int_0^1 (xe^{-y\tau}\sin (x\tau)+ye^{-y\tau}\cos x\tau)\,\mathrm{d}\tau\\ &=\Big[-e^{-y\tau}\cos(x\tau)\Big]_{\tau=0}^{\tau=1}=1-e^{-y}\cos x. \end{align*}