# Complexifying an integral.

1. Jul 3, 2011

### HotMintea

1. The problem statement

1.1. Is it possible to do $\int\ sin{x}\, \ cos{x}\, \ e^x\, \ dx\$ by complexifying the integral? (Note: not by integration by parts.)

Complexifying the Integral (Arthur Mattuck, MIT) [9:23]

1.2. When is it appropriate to complexify an integral, beside the condition that the integrand can be expressed as $Re (\ e^{\alpha x})\, \$?

2. The attempt at a solution

2.1.

$$\begin{equation*} \begin{split} \int\ sin{x}\, \ cos{x}\, \ e^{x}\, \ dx\ =\\ \int\ Re(\ e^{i(\frac{\pi}{2}\ -\ x)}\ )\, \ Re(\ e^{ix})\, \ e^{x}\ dx\ =\\ Re\int\ e^{i(\frac{\pi}{2}\ -\ x)}\, \ e^{ix}\, \ e^{x}\ dx\ =\\ Re\int\ i\ e^x\, \ dx\ =\\ - Im( e^x )\ + \ C\, \, (?) \end{split} \end{equation*}$$

Last edited by a moderator: Sep 25, 2014
2. Jul 3, 2011

### micromass

Hi HotMintea!

You've made a mistake in your solution. You seem to think that Re(z)Re(z')=Re(zz'), but this is not the case. The real part doesn't behave that way. Thus you cannot say

$$Re(e^{i(\pi/2-x)})Re(e^{ix})=Re(e^{i(\pi/2-x)+ix})$$

Instead, it would be better to use some trigonometric formula's in the beginning:

$$\sin(x)\cos(x)=\frac{\sin(2x)}{2}$$

3. Jul 3, 2011

### HotMintea

Hi micromass!

Thank you for pointing out my mistake and also for the suggestion.

$$\begin{equation*} \begin{split} \int\ sin{x}\, \ cos{x}\, \ e^x\ dx\, = \, \frac{1}{2}\int\ sin{2x}\, \ e^x\ dx\, \ = \\ \frac{1}{2}\, \ Re\int\ e^{i(\frac{\pi}{2} - 2x)}\, \ e^x\, \ dx\, = \, \ \frac{1}{2}\ Re(\frac{i}{1-2i}\ e^{(1-2i)x})\, \ + \, \ C\, \ = \\ \frac{1}{10}\ e^x\ Re((i-2)(cos{2x}-isin{2x}))\, +\, \ C\, \ = \, \frac{1}{10}\ e^x\, \ (sin{2x}-2cos{2x})\, \ + \, \ C \end{split} \end{equation*}$$

It now seems to me that complexification works and is effective if an integrand can be rewritten as $Re(e^{\alpha x}$). Am I correct?

4. Jul 3, 2011

5. Jul 3, 2011