# I need a hand with this integral? probably not even that difficult?

1. Oct 18, 2009

### jeebs

Hi,
I need to evaluate $$\int e\stackrel{-(r/R)}{}sin(qr/h)dr$$ between r=0 and r=$$\infty$$ but I keep getting stuck.

I hae tried integrating by parts but the integral just keeps getting more complicated, and I have tried substituting in sin(x) = (1/2i)(e^ix - e^-ix) but that gave me the result that the whole thing is equal to infinity, which I know to be wrong.

can anybody suggest some sort of method or substitution that I could use?
this is driving me up the wall now.
cheers.

2. Oct 18, 2009

### Donaldos

And yet it should have worked... Did you find an antiderivative for $$e^{-\frac r R}\sin\left(\frac{qr}{h}\right)$$ ?

3. Oct 18, 2009

### Staff: Mentor

Your first approach would have worked if you chose the right things for u and dv, and if you had done integration by parts a second time.

For the first integration by parts, choose u = e-Ar and dv = cos(Br)dr. For the sake of simplicity, A = 1/R and B = q/h.

That should give you an integral with e-Ar and sin(Br). Perform integration by parts a second time, with u = e-Ar and dv = sin(Br)dr. This second integration by parts will give you an integral with e-Ar and cos(Br), which is the same as what you started with. At this point you will have your original integral on the left side, and some expression plus a multiple of your original integral on the right side.

Bring both integral terms to one side and solve algebraically for your integral.

4. Oct 18, 2009

### tiny-tim

Hi jeebs!

(have an integral: ∫ and an infinity: ∞ and try using the X2 tag just above the Reply box )

Hint: what is e-r/Rsin(qr/h) the imaginary part of?

5. Oct 18, 2009

### HallsofIvy

Staff Emeritus
Let $u=e^{r/R}$ and $dv= sin(qr/h)dr$. Then $du= e^{r/R}/R dr$ and $v= -(h/q) cos(qr/h)$.

$\int e^{r/R}sin(qr/h)dr= -(h/q)e^{r/R}cos(qr/h)+ (h/(qR))\int e^{r/R}cos(qr/h)dr$. Now do it again, again letting $u= e^{r/R}$ and $dv= sin(qr/h)r$. You will get a term involving $\int e^{r/R}sin(qr/h)dr$ again. Do not integrate by parts a third time. Instead, solve the equation, algebraically, for $\int e^{r/R}sin(qr/h)dr$.