Integration involving complex numbers

[Useful nucleus]

Homework Statement

$$\int x* exp(-\alpha x -ikx)dx$$

the integration is definite with limits (x: 0---->inf)
alpha , k are real constants


2. The attempt at a solution

I used integration by parts and arrived at the result

$$\frac{xexp(-\alpha x-ikx)}{\alpha+ik}$$ + $$\frac{exp(-\alpha x-ikx)}{(\alpha+ik)^{2}}$$

now when i come to substitute by the integration limits comes the problem. How to deal with inf. ??
I know that the complex exponential functions are periodic , so there limit when x-->inf will never have certain value.

Any suggetions will be highly appreicated.

 

[chaoseverlasting]

If you know something about the gamma function, you could substitute (a+ik)x=t which would give you $$\frac{2!}{(a+ik)^2}$$ as the solution.

 

[Dick]

Homework Statement

$$\int x* exp(-\alpha x -ikx)dx$$

the integration is definite with limits (x: 0---->inf)
alpha , k are real constants2. The attempt at a solution

I used integration by parts and arrived at the result

$$\frac{xexp(-\alpha x-ikx)}{\alpha+ik}$$ + $$\frac{exp(-\alpha x-ikx)}{(\alpha+ik)^{2}}$$

now when i come to substitute by the integration limits comes the problem. How to deal with inf. ??
I know that the complex exponential functions are periodic , so there limit when x-->inf will never have certain value.

Any suggetions will be highly appreicated.

That's exp(-ax)*(cos(-kx)+i*sin(-kx)). If you assume (and you had better) that a>0 and k is real, then the exponential goes to zero at inf so the oscillatory part doesn't matter. You can also use l'Hopital's rule to find the limit of x*exp(-ax) as x->inf.

 

[Useful nucleus]

chaoseverlasting,

If you know something about the gamma function, you could substitute (a+ik)x=t which would give you as the solution

I know about Gamma function , but the reason that made me discard it is ,when you change the integration limits after choosing the substitution (a+ik)x=t , I have the following problem:
x goes to inf, then t goes to what?? (inf. +i inf. )!



Dick,

That's exp(-ax)*(cos(-kx)+i*sin(-kx)). If you assume (and you had better) that a>0 and k is real, then the exponential goes to zero at inf so the oscillatory part doesn't matter. You can also use l'Hopital's rule to find the limit of x*exp(-ax) as x->inf.

I believe your argument regarding the second term, which is exp(-ax) will go to zero when x goes to inf regardless of the behavior of the oscillatory part.
However, I'm still not convinced of using L'Hopital rule for the first term. Would you mind elaborating on that?


Anyways thank you chaoseverlasting and Dick for your help!

 

[Dick]

Sure. x*exp(-ax) is an infinity*0 form. Change it into x/exp(ax). Now it's infinity/infinity and you can apply l'Hopital. What do you get?

 

[Useful nucleus]

Lim$$_{x \rightarrow \infty}$$ $$\frac{x}{exp(\alpha x) * exp( ikx)}$$

At this point , by direct substituton we get

$$\frac{\infty}{\infty * exp( ik \infty) }$$

So, is it still possible to use L'Hopital ,taking into account the complex part.Why?

IF L'Hopital is possible here , I will end up with zero. But my question is why it was possible to use L'Hopital although we have that complex part.

 

[Dick]

Just apply l'Hopital to x*exp(-ax). If that goes to zero, then x*exp(-ax)*exp(ikx) also goes to zero, right?

 

[Useful nucleus]

Yep. That is convencing , it is zero.

Thank you for help!