Use Laplace transform (with respect tot)to calculate the integral

I=\int([\cos(tx)/(x^2+a^2)]dx t\geqslant 0

[IMG]file:///C:/Users/Jamie/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]

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- May 11th 2011, 05:45 AMsublim25Laplace transform to calculate an integral.
Use Laplace transform (with respect to

*t**)*to calculate the integral

I=\int([\cos(tx)/(x^2+a^2)]dx t\geqslant 0

[IMG]file:///C:/Users/Jamie/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG] - May 11th 2011, 05:54 AMTheEmptySet
- May 11th 2011, 06:02 AMsublim25
Limits are from 0 to infinity.

- May 11th 2011, 06:41 AMTheEmptySet
Let

$\displaystyle I(t) =\int_{0}^{\infty}\frac{\cos(tx)}{x^2+a^2}dx$

Then

$\displaystyle \mathcal{L}\{I\} =\int_{0}^{\infty}\frac{\mathcal{L}\{\cos(tx)\}}{x ^2+a^2}dx=\int_{0}^{\infty}\frac{s}{s^2+x^2}\cdot \frac{1}{x^2+a^2}dx$

Now by partial factions we get

$\displaystyle \mathcal{L}\{I\} =\frac{s}{s^2-a^2}\int_{0}^{\infty}\left( \frac{1}{x^2+a^2}-\frac{1}{x^2+s^2}\right)dx=\frac{\pi}{2}\left( \frac{s}{a(s^2-a^2)}-\frac{1}{s^2-a^2}\right)$

Now if you take the inverse Laplace transform we get

$\displaystyle I=\frac{\pi}{2a}\cosh(at)-\frac{\pi}{2}\sinh(at)$ - May 11th 2011, 06:45 AMsublim25
Can you please explain how you get the partial fractions to come out to this?

- May 11th 2011, 06:54 AMTheEmptySet
$\displaystyle \frac{s}{(x^2+a^2)(x^2+s^2)}=\frac{Ax+B}{x^2+a^2}+ \frac{Cx+D}{x^2+s^2}$

$\displaystyle s =(Ax+B)(x^2+s^2)+(Cx+D)(x^2+a^2)$

Now expand all of this out to get

$\displaystyle s=Ax^3+Bx^2+s^2Ax+s^2B + Cx^3+Dx^2+a^2Cx+a^2D $

$\displaystyle s=(A+C)x^3+(B+D)x^2+(s^2A+a^2C)x+(s^2B+a^2D)$

So this gives 4 equations in the 4 unknowns A,B,C,D

$\displaystyle A+C=0 \quad B+D=0 \quad s^2A+a^2C=0 \quad s^2B+a^2D=s$

dont forget that s and a are constants. Now just solve this system. - May 11th 2011, 07:24 AMsublim25
I am not sure what I am doing wrong, but when I solve the system, everything is cancelling out.

- May 11th 2011, 07:38 AMTheEmptySet