5
$\begingroup$

I'm trying to integrate $$ I = \int {e^x \sin (k \pi x)} dx. $$ I've used Matlab and Wolfram Alpha, which have both given me the result $$ I = \frac{e^x(\sin (k \pi x) - \cos (k \pi x))}{k^2 \pi^2 +1 }.$$

How did these programs get this result?

$\endgroup$
5
$\begingroup$

Consider $e^{ik\pi x} = \cos k\pi x + i \sin k\pi x$. The integrand becomes $e^{(ik\pi + 1)x}$, then take the imaginary part after integrating. We get:

$$ \int e^{(ik\pi + 1) x} dx=\frac{e^{(ik\pi+1)x}}{ik\pi + 1}+C = \frac{(1-ik\pi)e^{(ik\pi+1)x}}{1+k^2\pi^2}+C=\frac{(1-ik\pi)e^x (\cos k\pi x+ i \sin k\pi x)}{1+k^2\pi^2}+C $$

Take the imaginary parts of both sides, should yield same result as integration by parts.

$$ \therefore \int e^x \sin k\pi x \,dx =\frac{e^x (\sin k\pi x - k\pi \cos k\pi x)}{1+k^2\pi^2}+C $$

$\endgroup$
  • $\begingroup$ this ought to be taught in first calculus courses along with basic complex arithmetic. Nice answer. $\endgroup$ – James S. Cook Oct 1 '14 at 4:35
3
$\begingroup$

$$I = \int e^x \sin (k \pi x) dx\\=\sin k\pi x\int e^xdx-\int k\pi\cos k\pi x \left(\int e^xdx \right)dx\\=e^x\sin k\pi x-k\pi\int e^x\cos k \pi x\\=e^x\sin k\pi x-k\pi\left(\cos k\pi x\int e^x dx-\int k\pi \sin k\pi x\left(\int e^x dx\right)dx\right) \\ I=e^x(\sin k \pi x -k\pi\cos k\pi x)+k^2\pi^2 I+c$$ $$I=\frac{e^x(\sin k\pi x-k\pi\cos k\pi x)}{1+k^2\pi^2}+c$$


In general: $$I_{a,b}=\int e^{ax}\sin bx dx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+c$$

$\endgroup$
  • $\begingroup$ Also a good solution! @James Harrison's jsut looked nicer. $\endgroup$ – jamesh625 Oct 1 '14 at 14:23
2
$\begingroup$

Hint: Use integration by parts twice.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.