Integral OH YES!

1. May 9, 2005

ziddy83

Ok, here is the integral i seem to be having some issues with. I know there's a very simple step im missing.

$$\int_{}^{} e^x \sin(\pi x) dx$$

i attempted to do this using by parts integration.

I tried u = $$\sin(\pi x)$$ so du= $$\pi \cos(\pi x) dx$$
so then dv= $$e^x dx$$ and v= $$e^x$$

after using $$uv- \int v du$$ I seem to be going in circles...can someone help?

2. May 9, 2005

quasar987

Do integration by parts on $\int v du$. You'll end up with

$$\int_{}^{} e^x \sin(\pi x) dx = something - \int_{}^{} e^x \sin(\pi x) dx$$

Aaah.. add $\int_{}^{} e^x \sin(\pi x) dx$ on both sides. Divide by two. ta-dam.

3. May 9, 2005

shmoe

Small warning, it won't look quite like you've described, the constant won't be two.

4. May 10, 2005

quasar987

Oh right.. because of the pies!

5. May 10, 2005

Galileo

There's a quick alternative way using complex exponentials.

Since
$$e^x \sin (\pi x)= \Im (e^{x+i\pi x})=\Im (e^{x(1+i\pi)})$$

$$\int e^{x(1+i\pi)} dx=\frac{e^{x(1+i\pi)}}{1+i\pi}=\frac{(1-i\pi)e^{x(1+i\pi)}}{1+\pi^2}$$

Now take the imaginary part.