# Integration variable is Imaginary and real d(Imx) , d(Rex)

So i have seen things like $$\int ... d(\cos x)$$ and now i have an expression with $$\int d\operatorname{Im}(z)$$ so i though my integral would be quite straight forward, knowing the Imaginary part of $$z$$.

$$z = \rho \, e^{i \theta}$$

This is my Integral in full: $$$$\int_{-\infty}^{+\infty} d\operatorname{Im}(z) \int_{-\infty}^{+\infty} d\operatorname{Re}(z) \; e^{-|z|^2} \; (z^*)^n \; z^m \label{imre-eqn} \tag{1}$$$$

This is the next step, which i cant get to: $$$$\int_{0}^{+\infty} d\rho \; \rho^{n+m+1} \; e^{-\rho^2} \; \int_{0}^{2\pi} d\theta \; e^{i\theta (m-n)} \label{exp-eqn} \tag{2}$$$$

The problem is basically that i don't know where to search to find how to get from (\ref{imre-eqn}) to (\ref{exp-eqn}). any tipp? or what can i put into google to find out ?

Or, can somebody give me a more simple example of changing the integral variable from Im and Re to $$\to$$ what looks like spherical coordinates

that would be fantastic thankyou!

Thanks! xyz

You can use the identities $$\Im(z)=-\frac{i}{2} (z-\bar{z})$$ and $$\Re(z)=\frac{1}{2}(z+\bar{z})$$. You will see that $$\Im(z)=\rho\sin(\theta)$$ and $$\Re(z)=\rho\cos(\theta)$$. It should be easy from here to transform your coordinates to $$(\rho,\theta)$$, I hope this helps.

• I have put d(ρ cos( θ)) into the integral d- part but don't know how to resolve it.. Why is the solution just d θ and dρ? Ι am German and don't know how to google it in English :( how do I convert to spherical Koordinates.. Thanks for your answere x
– Ella
Commented Dec 17, 2021 at 23:57
• There is a typo here. $\Im(z)=\rho\sin\theta$. Welcome Mr. Calculus; some MathJax advice - use backslashes to get special functions! If you're unsure how to render something, chances are that placing a backslash together with a sensible abbreviation will work. E.g., \Im, \rho, \sin\theta all render properly Commented Dec 18, 2021 at 0:13
• Op is a struggling with the transformation itself - worth fleshing out the answer a bit more Commented Dec 18, 2021 at 0:14
• I'm getting $\rho^{n+m\color{red}{+}1}$ and $e^{\color{red}{+}\rho^2}$ in my derivation, but I could be wrong. As for backslashes I just meant that Mr Calculus is writing im = rho sin theta when they should write \Im(z)=\rho\sin\theta @Ella Commented Dec 18, 2021 at 0:28
• Right. How do I get to the $/rho ^ {n+n+1}$ how do I split the d{/rho *cos / theta "}? Just into two integrals? d\$/rho} d{cos/theta"}
– Ella
Commented Dec 18, 2021 at 0:41

all i needed was someone to remind me that Re and Im can be drawn just like x and y in a koordinate system! And so the dIm and dRe can be translated to polar coordinates!