1. ## Need help with an integral please

$\displaystyle \int \frac{N}{2}\sigma\sin{(\omega \ln{\sigma })}+\frac{N}{2}\sigma\; d\sigma$

This is how far I've gotten...

$\displaystyle \frac{N}{2}\int \sigma\sin{(\omega \ln{\sigma })}+\sigma\; d\sigma$

$\displaystyle u\equiv \sin(\omega \ln{\sigma})$

$\displaystyle du = \frac{\omega \cos(\omega \ln{\sigma})}{\sigma}\; d\sigma$

$\displaystyle d\sigma = \frac{\sigma}{\omega \cos(\omega \ln{\sigma})}\; du$

$\displaystyle \frac{N}{2}\int \frac{u\sigma^2}{\omega \cos(\omega \ln{\sigma})}\; du +\frac{N}{2}\int \sigma\; d\sigma$

...not sure what to do...sos...sos...

2. Originally Posted by rainer

$\displaystyle \int \frac{N}{2}\sigma\sin{(\omega \ln{\sigma })}+\frac{N}{2}\; d\sigma$

This is how far I've gotten...

$\displaystyle \frac{N}{2}\int \sigma\sin{(\omega \ln{\sigma })}+1\; d\sigma$

$\displaystyle u\equiv \sin(\omega \ln{\sigma})$

$\displaystyle du = \frac{\omega \cos(\omega \ln{\sigma})}{\sigma}\; d\sigma$

$\displaystyle d\sigma = \frac{\sigma}{\omega \cos(\omega \ln{\sigma})}\; du$

$\displaystyle \frac{N}{2}\int \frac{u\sigma^2}{\omega \cos(\omega \ln{\sigma})}\; du +\frac{N}{2}\int 1\; d\sigma$

...not sure what to do...sos...sos...

There are three methods : the first one also the most effective one i suggest is sub.$\displaystyle \ln{\sigma} = t$ or $\displaystyle \sigma = e^t$

$\displaystyle d\sigma = e^t~dt$ so we have

$\displaystyle \frac{N}{2} \int (e^t)\sin(\omega t) (e^t~dt)$

$\displaystyle = \frac{N}{2} \int e^{2t} \sin(\omega t )~dt$

Definitely , it is an integral which is more familiar to you , you can solve it yourself now

The second method is using integration by parts , you can discover how it can work .

The third is the most powerful but it requires much attention :

$\displaystyle \int x\sin(\omega\ln{x})~dx$

$\displaystyle = lm\left\{\ \int x e^{i \omega \ln{x} }~dx \right\}\$

$\displaystyle = lm\left\{\ \int x e^{ \ln{x ^{i \omega }} }~dx \right\}\$

$\displaystyle = lm\left\{\ \int x \cdot x^{i \omega }~dx \right\}\$

$\displaystyle = lm\left\{\ \int x^{1+ i \omega }~dx \right\}\$

$\displaystyle = lm[ \frac{x^{2+i\omega} } { 2+i\omega} ] + C$

$\displaystyle = lm[ \frac{x^2 exp \{\ \ln{ x^{i\omega}} \}\ ( 2 - i\omega )}{ 4 + \omega^2 } ] + C$

$\displaystyle = \frac{ x^2 ( 2\sin(\omega\ln{x}) - \omega \cos(\omega\ln{x}))}{ 4 + \omega^2 } + C$

3. Originally Posted by rainer

$\displaystyle \int \frac{N}{2}\sigma\sin{(\omega \ln{\sigma })}+\frac{N}{2}\sigma\; d\sigma$

This is how far I've gotten...

$\displaystyle \frac{N}{2}\int \sigma\sin{(\omega \ln{\sigma })}+\sigma\; d\sigma$

$\displaystyle u\equiv \sin(\omega \ln{\sigma})$

$\displaystyle du = \frac{\omega \cos(\omega \ln{\sigma})}{\sigma}\; d\sigma$

$\displaystyle d\sigma = \frac{\sigma}{\omega \cos(\omega \ln{\sigma})}\; du$

$\displaystyle \frac{N}{2}\int \frac{u\sigma^2}{\omega \cos(\omega \ln{\sigma})}\; du +\frac{N}{2}\int \sigma\; d\sigma$

...not sure what to do...sos...sos...
http://www.wolframalpha.com/input/?i...5D+%2B+N%2F2*x

Click Show Steps

4. Originally Posted by rainer

$\displaystyle \int \frac{N}{2}\sigma\sin{(\omega \ln{\sigma })}+\frac{N}{2}\; d\sigma$

This is how far I've gotten...

$\displaystyle \frac{N}{2}\int \sigma\sin{(\omega \ln{\sigma })}+1\; d\sigma$
$\displaystyle u \equiv \ln{\sigma}$

$\displaystyle du = 1/\sigma d\sigma$ implies $\displaystyle d\sigma = \sigma du$

$\displaystyle \frac{N}{2} \int e^{2u} \sin(\omega u)du + \frac{N}{2}\int 1\; d\sigma$

$\displaystyle \frac{Ne^{2u}\sin(\omega u)}{4} + \frac{N}{2} \int e^{2u} \omega \cos(\omega u)du + \frac{N\sigma}{2}$

$\displaystyle \frac{N\sigma ^2}{4}(\sin (\omega \ln{\sigma})+ \omega \cos (\omega \ln{\sigma}))- \frac{N\omega ^2}{2}\int e^{2u}\sin(\omega u)du ~+$ $\displaystyle \frac{N\sigma}{2} = \frac{N}{2} \int e^{2u} \sin(\omega u)du + \frac{N\sigma}{2}$

$\displaystyle \frac{N}{2}(1+\omega ^2) \int e^{2u} \sin(\omega u)du = \frac{N\sigma ^2}{4}(\sin (\omega \ln{\sigma})+ \omega \cos (\omega \ln{\sigma}))$

$\displaystyle \int \sigma\sin{(\omega \ln{\sigma })} d\sigma = \frac{\sigma ^2(\sin (\omega \ln{\sigma})+ \omega \cos (\omega \ln{\sigma}))}{2+2\omega ^2}$

$\displaystyle \frac{N}{2}\int \sigma\sin{(\omega \ln{\sigma })}+1\; d\sigma = \frac{N \sigma ^2(\sin (\omega \ln{\sigma})+ \omega \cos (\omega \ln{\sigma}))}{4+4\omega ^2} + \frac{N\sigma}{2} + C$

...which doesn't agree with mathematica. I should not be allowed to answer integral questions at 1 am :/

5. Originally Posted by simplependulum
$\displaystyle = \frac{N}{2} \int e^{2t} \sin(\omega t )~dt$

Definitely , it is an integral which is more familiar to you , you can solve it yourself now
Oh yes, yes indeed.

Originally Posted by simplependulum
The third is the most powerful but it requires much attention
Over my head for now, but very interesting. At first I did the integral by substituting euler's relation for the sine function. That made the integral very easy going, but the result was totally useless to me. "i" and me don't get along very well.

Originally Posted by Turiski
...which doesn't agree with mathematica. I should not be allowed to answer integral questions at 1 am :/
Turiski- The magic recipe for integrating $\displaystyle \int e^{2t} \sin(\omega t )~dt$ has a minus sign where you have a plus sign. It also has the exponential factor 2 down in front of sin(wln(s)). Otherwise looks like you were on target.

Thanks everybody!