# Need to find the integral of the following function using Laplace Transform int_0^(oo) (cos(6t)-cos(4t))/(t)dt

Need to find the integral of the following function using Laplace Transform
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(6t\right)-\mathrm{cos}\left(4t\right)}{t}\phantom{\rule{thinmathspace}{0ex}}dt$
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Baluttor7
Hint : write it as
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}6t-\mathrm{cos}4t}{t}{e}^{-st}dt$
Where s= 0
Now it become Laplace transformation
$L\left(\frac{\mathrm{cos}6t-\mathrm{cos}4t}{t}\right)$
We have property
$L\left(\frac{f\left(t\right)}{t}\right)={\int }_{s}^{\mathrm{\infty }}F\left(s\right)ds$
Here $F\left(s\right)=L\left(f\left(t\right)\right)$
Using above property
$L\left(\frac{\mathrm{cos}6t-\mathrm{cos}4t}{t}\right)={\int }_{s}^{\mathrm{\infty }}\left(\frac{s}{{s}^{2}+{6}^{2}}-\frac{s}{{s}^{2}+{4}^{2}}\right)ds$
Integrate you will get
$=\frac{1}{2}\mathrm{ln}\left(\frac{{s}^{2}+{4}^{2}}{{s}^{2}+{6}^{2}}\right)$
Now put s=0
=
$=\frac{1}{2}\mathrm{ln}\left(\frac{{4}^{2}}{{6}^{2}}\right)$
$=\mathrm{ln}\left(2/3\right)$