Solving Integral Using Laplace Transform

[pierce15]

I posted this in the homework section, but I haven't received any help, so hopefully putting it in this section won't be an issue. I'm trying to compute the integral

$$ \int_0^ \infty \frac{ \cos xt}{1 + t^2} \, dt $$

using the Laplace transform. The first thing that catches my eye is the 1 /(1 + t^2) factor, which is equal to the Laplace transform of sin x:

$$ = \int_0^\infty \cos xt \, L[ \sin x ] \, dt $$

Any ideas?

 

[vanhees71]

Why do you want to use the Laplace transformation here? I'd rather use the Fourier transformation. It's simpler to put it first in the exponential form. Your Integral is
$$F(x)=\frac{1}{2}\int_0^{\infty} \frac{\exp(\mathrm{i} x t)+\exp(-\mathrm{i} x t)}{1+t^2}.$$
Substituting $t'=-t$ in the second integral you get after some algebra
$$F(x)=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\exp(\mathrm{i} x t)}{1+t^2}. \qquad (1)$$
Now we use the fact that
$$\int_{-\infty}^{\infty} \frac{\mathrm{d} x}{2 \pi} \exp(-|x|) \exp(-\mathrm{i} t x)=\frac{1}{\pi (1+t^2)}.$$
Thus (1) is (up to a factor $\pi/2$) the inverse of this Fourier transform. This gives
$$F(x)=\frac{\pi \exp(-|x|)}{2}.$$

 

[pierce15]

Ok, thanks for that. Do you also see any way to do it with the Laplace transform?

 

[vanhees71]

Hm, I've no idea. Perhaps you can somehow use the convolution theorem with some clever trick?

 

[blue_raver22]

The Laplace transform is a powerful tool in mathematics and engineering that allows us to convert functions from the time domain to the frequency domain. In this case, we are using it to solve an integral in the time domain by converting it to a simpler form in the frequency domain.

As you correctly pointed out, the 1/(1+t^2) factor is equal to the Laplace transform of sin x. This means that we can rewrite the integral as:

$$ = \int_0^\infty \cos xt \, L[ \sin x ] \, dt = \int_0^\infty \cos xt \, \frac{1}{s^2+1} \, dt $$

where s is the complex frequency variable.

Next, we can use the Laplace transform property of convolution to rewrite the integral as:

$$ = \frac{1}{2\pi i} \int_{c-i \infty}^{c+i \infty} \frac{1}{s^2+1} \, L[\cos xt] \, ds $$

where c is a constant chosen such that the integral converges.

Using the Laplace transform of cosine, we get:

$$ = \frac{1}{2\pi i} \int_{c-i \infty}^{c+i \infty} \frac{1}{s^2+1} \, \frac{s}{s^2+x^2} \, ds $$

Finally, we can use the residue theorem to evaluate this integral and obtain the solution:

$$ = \frac{1}{2}e^{-x} $$

In summary, by using the Laplace transform and its properties, we were able to solve the given integral in a more efficient and elegant way. This highlights the usefulness of the Laplace transform in solving various mathematical problems.