Question

An integral a day:
Day two, this one has a nifty trick to it, I'll post a full solution in an hour or so.

\[\int_0^\infty sin(ax)e^{-sx}dx \]

Answer 1

what's in the exponent? s?

Answer 2

ok the laplace, I get it
hold on, I can do this...

Answer 3

first integrate by parts I believe\[\lim_{n\to\infty}\int_{0}^{n} e^{-st}\cos(at)dt\]\[=\lim_{n\to\infty}\frac1ae^{-st}\sin(at)|_{0}^{n}-\frac s{a^2}e^{-st}\cos(at)|_{0}^{n}-\frac{s^2}{a^2}\int_{0}^{n} e^{-st}\cos(at)dt\]hopefully evaluating this works out...

Answer 4

the first evaluation is\[\lim_{n\to\infty}\frac1ae^{-sn}\sin(an)=0\]the next term evaluates to\[\lim_{n\to\infty}\frac s{a^2}(e^{-sn}\cos(an)-1)=-\frac s{a^2}\]so we get\[I=\frac s{a^2}-\frac{s^2}{a^2}I\implies(1+\frac{s^2}{a^2})I=\frac s{a^2}\]now I can only hope the algebra works out...

Answer 5

and it does :)\[I=\frac s{a^2}({a^2\over s^2+a^2})={s\over s^2+a^2}=\int_{0}^{\infty}e^{-st}\cos(at)dt=\mathcal L\{\cos(at)\}\]

Answer 6

wow I did the whole thing for cosine, I need to read more carefully
oh well, sin is the same deal

Answer 7

for yours, after parts you get\[I=-\frac1ae^{-sx}\cos(ax)|_{0}^{\infty}-\frac s{a^2}e^{-sx}\sin(ax)|_{0}^{\infty}+\frac{s^2}{a^2}I\]take the limits and evaluate and you get\[I=\frac1a-\frac{s^2}{a^2}I\]solve for I

I am scared because James is typing, which makes me think I made a mistake..

Answer 8

Alternatively, write
\[ \sin(ax) = \frac{1}{2i} ( e^{aix} - e^{-iax} ) \]
Then
\[ \sin(ax)e^{-sx} = \frac{1}{2i}( e^{(-s+ai)x} - e^{-(s+ai)x} ) \]
which has definite integral
\[ -\frac{1}{2i}\left( \frac{1}{-s + ai} + \frac{1}{s + ai} \right) \]
and simplifies to
\[ -\frac{1}{2i}\cdot \frac{-2ai}{s^2 + a^2} = \frac{a}{s^2+a^2} \]

Answer 9

...but I haven't studied complex analysis :(

Answer 10

oh I guess that's not really

Answer 11

...just euler's formula

Answer 12

This doesn't really need complex analysis. Just a suspension of doubt that using complex coefficients will lead you awry.

Answer 13

Thanks, I often need reassurance that complex numbers are useful, haha

Answer 14

*correction: (sigh) there's a sign error in my working, but the final answer is correct!

Answer 15

Good job to those who figured it out, this is how I would have done it:
Let the integral equal to C:
\[C= \int_0^\infty \sin(ax)e^{-sx}dx \\ \text{Integrate by parts twice:} \\ C = \frac{-\sin(ax)e^{-sx}}{s} |_0^\infty + \int_0^\infty \frac{a \cos(ax)e^{-sx}}{s}dx\]\[C=\frac{a\cos(ax)e^{-sx}}{s^2}|_0^\infty - \int_0^\infty\frac{a^2\sin(ax) e^{-sx}}{s^2}dx\]\[C =\frac{a}{s^2} -\frac{a^2}{s^2}C\]\[(1+\frac{a^2}{s^2})C = \frac{a}{s^2}\]\[C =\frac{a}{s(1+\frac{a^2}{s^2})}=\frac{a}{s^2+a^2}\]

Answer 16

And that's exactly the same as TuringTest's solution, right on.