| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Can anyone point me to the identities etc. that are used to come up
with the solution to the following Integral?
(Sum = Integral sign)
ax
ax e
Sum e Sin bx dx = -------- ( a Sin bx - b Cos bx )
2 2
a + b
I have tried to work out the solution and keep ending up down a rat
hole. I have tried converting the integral to all exponentials and
that didn't work for me either.
I am doing a Fourier series solution where a = -j*pi*(2n-1) and
b = -j*pi*(2n+1). Just for fun I am trying to put all this in the
integral and work the solution through long hand, rather than look-up
table.
Any ideas?
Dennis
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1495.1 | by parts | STAR::ABBASI | Fri Sep 20 1991 12:37 | 42 | |
by parts:
I= sum e^ax sin bx
I= e^ax cos bx - SUM (cos bx)a e^ax
----------- ---------------
b b
I= e^ax cos bx - a/b ( -e^ax sin bx + SUM ( sin bx a e^ax) )
----------- ------------ -----------------
b b b
I= e^ax cos bx - a/b ( -e^ax sin bx + a/b I )
----------- ------------
b b
I = e^ax cos bx - a e^ax sin bx - a^2 I
------------ ------------- ---
b b^2 b^2
I + a^2 I = e^ax ( b cos bx - a sin bx)
---- ---------------------------
b^2 b^2
I ( b^2 +a^2 ) = "
-----------
b^2
I= e^ax ( b cos bs - a sin bx) QED
---------------------
b^2 + a ^2
also if you expand sin in exponentials (e^bi - e^-bi )/2i
it will work.
/nasser
| |||||
| 1495.2 | ZFC::deramo | the radio reminds me | Fri Sep 20 1991 20:31 | 9 | |
Or just take e^ax sin bx to be the coefficient of i in the imaginary part of e^ax e^ibx = e^(a+ib)x. The indefinite integral becomes e^(a+ib)x / (a+ib) which equals (a-ib)e^(a+ib)x / (a^2 + b^2), and the coefficient of i in that is ( e^ax / (a^2 + b^2) )( a sin bx - b cos bx ). This is essentially the second suggestion in .1. Dan | |||||