Efficient Techniques for Solving Integrals: Quick and Simple Approach

[Quinzio]

Homework Statement

I'm trying to solve an integral, but I can't reach the book result.
I've done my work, I filled 4 sheets with integrals, but somewhere I make a mistake.

I just ask, if there a quick short way to solve the integral, to tell me or show me.

\int_{0}^{x} (x-t)e^{m(x-t)}cos(at) dtI have another question...
is this
\int_{0}^{x}du\int_{0}^{u} e^{m(x-t)}cos(at) dt

equivalent to ...

\int_{0}^{x}\int_{0}^{u} e^{m(x-t)}cos(at) dt du

Can I move the variable in an out the integral ?
Is the first form just a way to explicit which varibale belongs to which integration bounds ?

Homework Equations

NA

The Attempt at a Solution

NA

 

[losiu99]

To avoid complications consider \int t e^t \cos t \dt and integrate by parts (u = t,\mbox{d}v=e^t \cos t\, \mbox{d} t). The result is awful, though.

 

[jackmell]

I have another question...
is this
\int_{0}^{x}du\int_{0}^{u} e^{m(x-t)}cos(at) dt

equivalent to ...

\int_{0}^{x}\int_{0}^{u} e^{m(x-t)}cos(at) dt du

Can I move the variable in an out the integral ?
Is the first form just a way to explicit which varibale belongs to which integration bounds ?

I'm no analyst but I would say yes. They are exactly the same and just different ways of expressing the double integral: you have an inner integral function f(u). The outer integral then integrates that function over it's bounds. Note if you switched the order of integration, say to dudt, then some changes to the limits would usually be needed. Left that to the interested reader.

 

[vela]

One thing you might try is using the fact that

(x-t)e^{m(x-t)} = \frac{\partial}{\partial m} e^{m(x-t)}

so that

\int_{0}^{x} (x-t)e^{m(x-t)}\cos(at)\,dt = \int_{0}^{x} \frac{\partial}{\partial m}e^{m(x-t)}\cos(at)\,dt = \frac{\partial}{\partial m}\int_{0}^{x} e^{m(x-t)}\cos(at)\,dt

Also, use e^m(x-t) = e^mxe^-mt and pull the factor e^mx out of the integral.

 

[fzero]

I would use Euler's identity and solve for cos:

\cos \alpha = \frac{e^{i\alpha} + e^{-i\alpha}}{2}.

The resulting integrals are easily done by parts.

As for the second question about moving differentials, it's find as long as you do the integration in the appropriate order and don't confuse limits.