Is Complexifying an Integral the Solution for Tricky Integrands?

[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.

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

 

[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:

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

 

[HotMintea]

Hi micromass!

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

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

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?

 

[micromass]

Hi micromass!

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

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

In the very last step, you should write 1/5 instead of 1/10. For the rest you are correct.

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?

Yes, that's true!

 

[HotMintea]

In the very last step, you should write 1/5 instead of 1/10. For the rest you are correct.

1/10 seems correct: http://www.wolframalpha.com/input/?i=int+cos%28x%29sin%28x%29e^x+dx&asynchronous=false&equal=Submit

Thanks again for your help!