Evaluating an integral with Laplace

Question

We need to evaluate the following integral:

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

There is the following note: "You may interchange taking the Laplace transform and integrating."

I have no idea how to solve this, I need help. Thank you!

Answer 1

Let $$f(t)=\int_0^{\infty} \frac{cos(tx)}{x^2+a^2}\, \mathrm{d}x$$

Then, $$\mathcal{L} (f(t)) = \int_0^{\infty} e^{-st}\left( \int_0^{\infty} \frac{cos(tx)}{x^2+a^2}\, \mathrm{d}x\right) \, \mathrm{d}t$$

On interchanging, $$\mathcal{L} (f(t)) = \int_0^{\infty} \frac{1}{x^2+a^2} \left( \int_0^{\infty} e^{-st} cos(tx) \, \mathrm{d}t \right) \, \mathrm{d}x$$

Now utilizing the Laplace Transform of $cos(tx)$ to solve the inner integral, we have $$\mathcal{L} (f(t)) = \int_0^{\infty} \frac{s}{(x^2+a^2)(x^2+s^2)} \, \mathrm{d} x$$

On solving and simplifying this integral using partial fractions and the basic rules of definite integration, we have $$\mathcal{L} (f(t)) = \frac{\pi}{2a(a+s)}$$

Applying the inverse Laplace Transform on both sides, we have  $$f(t)=\mathcal{L}^{-1}\left(\frac{\pi}{2a(a+s)}\right)$$

$$f(t)=\frac{\pi}{2a} \mathcal{L}^{-1}\left(\frac{1}{a+s}\right)$$

Thus, $$f(t)=\frac{\pi}{2a} e^{-at}$$ and we are done.