# Quick way to solve this ?

## 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.

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

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 ?

NA

## The Attempt at a Solution

NA

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.

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
Staff Emeritus
Homework Helper
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 em(x-t) = emxe-mt and pull the factor emx out of the integral.

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