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?

3 Answers
Jun 5, 2018

See Explanation

Explanation:

#f(x)=\sum_(n=0)^oo(-ax)^n/(n!)(-1)^n((bx)^(2n+1))/((2n+1)!)#
#...=\sum_0^oo((-a)^nx^n)/(n!)(-1)^n(b^(2n+1)x^(2n+1))/((2n+1)!)#
#...=\sum_0^oo((-a)^nb^(2n+1))/(n!(2n+1)!)(-1)^n\int_0^\inftyx^nx^(2n+1)dx...#

Jun 5, 2018

# 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) #

Jun 5, 2018

#int_0^oo e^(-ax)sin(bx)dx = b/(a^2+b^2)#

Explanation:

Alternatively, using complex numbers:

#int_0^oo e^(-ax)sin(bx)dx = int_0^oo e^(-ax)(e^(ibx)-e^(-ibx))/(2i)dx#

#int_0^oo e^(-ax)sin(bx)dx = 1/(2i)int_0^oo e^(-ax+ibx)dx-1/(2i)int e^(-a-ibx)dx#

#int_0^oo e^(-ax)sin(bx)dx = 1/(2i)[ e^(-ax+ibx)/(-a+ib) +e^(-ax-ibx)/(a+ib)]_0^oo#

#int_0^oo e^(-ax)sin(bx)dx = 1/(2i)[ e^(-ax)(e^(ibx)/(-a+ib) +e^(-ibx)/(a+ib))]_0^oo#

#int_0^oo e^(-ax)sin(bx)dx = -1/(2i)[ e^(-ax)((a+ib)e^(ibx)+(-a+ib) e^(-ibx))/((a-ib)(a+ib))]_0^oo#

#int_0^oo e^(-ax)sin(bx)dx = -1/(2i)[ e^(-ax)(ae^(ibx)+ibe^(ibx)-ae^(-ibx)+ ibe^(-ibx))/(a^2+b^2)]_0^oo#

#int_0^oo e^(-ax)sin(bx)dx = -1/(2i)[ e^(-ax)/(a^2+b^2)(a(e^(ibx)-e^(-ibx))+ib(e^(ibx)+e^(-ibx)))]_0^oo#

#int_0^oo e^(-ax)sin(bx)dx = -[ e^(-ax)/(a^2+b^2)(a((e^(ibx)-e^(-ibx)))/(2i)+b((e^(ibx)+e^(-ibx)))/2)]_0^oo#

#int_0^oo e^(-ax)sin(bx)dx = -[ e^(-ax)/(a^2+b^2)(asin(bx)+bcos(bx))]_0^oo#

#int_0^oo e^(-ax)sin(bx)dx = b/(a^2+b^2)#