Find #\int_0^ooe^(-ax)\sin(bx)dx#?
I know these:
#e^-(ax)=\sum_(n=0)^oo((-ax)^n)/(n!)#
#\sin(bx)=\sum_(n=0)^oo(-1)^n((bx)^(2n+1))/((2n+1)!)#
However... (see my answer below) can I do that?
I know these:
#e^-(ax)=\sum_(n=0)^oo((-ax)^n)/(n!)# #\sin(bx)=\sum_(n=0)^oo(-1)^n((bx)^(2n+1))/((2n+1)!)#
However... (see my answer below) can I do that?
3 Answers
See Explanation
Explanation:
# int_0^oo \ e^(-ax) \ sin(bx) \ dx = b/(a^2+b^2)#
Explanation:
We seek:
# I = int_0^oo \ e^(-ax) \ sin(bx) \ dx #
We can apply Integration By Parts:
Let
# { (u,=e^(-ax), => (du)/dx,=-ae^(-ax)), ((dv)/dx,=sin(bx), => v,=-1/bcos(bx) ) :}#
Then plugging into the IBP formula:
# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #
We have:
# int_0^oo \ (e^(-ax))(sin(bx)) \ dx = [(e^(-ax))(-1/bcos(bx))]_0^oo - int_0^oo \ (-1/bcos(bx))(-ae^(-ax)) \ dx #
# :. I = -1/b \ [ \ e^(-ax) \ cos(bx) \ ]_0^oo - a/b \ int_0^oo \ e^(-ax) \ cos(bx) \ dx #
Now, Consider the second integral, For which we apply a second application of Integration By Parts:
Let
# { (u,=e^(-ax), => (du)/dx,=-ae^(-ax)), ((dv)/dx,=cos(bx), => v,=1/bsin(bx) ) :}#
So that:
# int_o^oo \ (e^(-ax))(cos(bx)) \ dx = [ \ (e^(-ax))(1/bsin(bx)) \ ]_o^oo - int_o^oo \ (1/bsin(bx))(-ae^(-ax)) \ dx #
# :. int_o^oo \ e^(-ax) \ cos(bx) \ dx = 1/b \ [ \ e^(-ax) \ sin(bx) \ ]_o^oo +a/b \ I #
And, Combining these results, we get:
# I = -1/b \ [ \ e^(-ax) \ cos(bx) \ ]_0^oo - a/b {1/b \ [ \ e^(-ax) \ sin(bx) \ ]_o^oo +a/b \ I} #
# :. I = -1/b \ [ \ e^(-ax) \ cos(bx) \ ]_0^oo - a/b^2 \ [ \ e^(-ax) \ sin(bx) \ ]_o^oo - a^2/b^2 \ I #
# :. b^2I = -b \ [ \ e^(-ax) \ cos(bx) \ ]_0^oo - a \ [ \ e^(-ax) \ sin(bx) \ ]_o^oo - a^2 \ I #
# :. (a^2+b^2)I = -b \ L_1 - a \ L_2 #
Where:
# L_1 = [ e^(-ax) \ cos(bx) ]_0^oo #
# \ \ \ \ = (lim_(x rarr oo) e^(-ax) \ cos(bx)) - (e^0cos0) #
# \ \ \ \ = (0) - (1) #
# \ \ \ \ = -1 #
And:
# L_2 = [ \ e^(-ax) \ sin(bx) \ ]_o^oo #
# \ \ \ \ = (lim_(x rarr oo) e^(-ax) \ sin(bx)) - (e^0sin0) #
# \ \ \ \ = (0) - (0) #
# \ \ \ \ = 0 #
So that:
# (a^2+b^2)I = b => I = b/(a^2+b^2) #
Explanation:
Alternatively, using complex numbers: