Struggling with Integration: e^x \sin(\pi x)

[ziddy83]

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

 

[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.

 

[shmoe]

Divide by two.

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

 

[quasar987]

Oh right.. because of the pies!

 

[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.