How to solve following integral using Laplace Transform

Question

I am looking for a closed form expression for the following integral. One approach that comes to my mind was to use Laplace transform, using the following, I came with the function $f(t)$. What can be the Laplace transform for this function.

$$\int^\infty_0f(r) dr = A\int^\infty_0 \frac{r}{1+Cr^4} e^{-Br^2} dr$$

If I consider $t= r^2$ and $s = B$

$$\int^\infty_0f(r) dr = \frac{A}{2}\int^\infty_0 \frac{1}{1+Ct^2} e^{-st} dt$$

This means now $$f(t) = \frac{1}{1 + Ct^2}$$ I cannot find $\mathcal{L}_{f(t)} (s)$ since it is again complicated. Any assistance would be highly appreciated.

PS: I am concerned with the closed form expression of the integral, if I can get it without Laplace transform, that is suffice for me. I will appreciate any help.

Answer 1

Here is a way to find the integral $$I = \int_0^\infty \frac{e^{-at}}{1 + bt^2} \, dt,$$ using the following result for Laplace transforms $$\int_0^\infty f(t) g(t) \, dt = \int_0^\infty \mathcal{L}\{f(t)\}(s) \cdot \mathcal{L}^{-1} \{g(t)\}(s) \, ds.$$ Setting $f(t) = e^{-at}$ where $a > 0$ and $g(t) = 1/(1 + b t^2)$ where $b > 0$, since $$\mathcal{L}\{e^{-at}\} = \frac{1}{a + s} \quad \text{and} \quad \mathcal{L}^{-1} \left \{\frac{1}{1 + b t^2} \right \} = \frac{1}{\sqrt{b}} \sin \left (\frac{s}{\sqrt{b}} \right ),$$ the integral can be rewritten as $$I = \frac{1}{\sqrt{b}} \int_0^\infty \frac{\sin (s/\sqrt{b})}{a + s} \, ds.$$ Enforcing the substitution of $s \mapsto s - a$ yields \begin{align*} I &= \frac{1}{\sqrt{b}} \int_a^\infty \sin \left (\frac{s - a}{\sqrt{b}} \right ) \frac{ds}{s}\\ &= \frac{1}{\sqrt{b}} \int_a^\infty \left [\sin \left (\frac{s}{\sqrt{b}} \right ) \cos \left (\frac{a}{\sqrt{b}} \right ) - \cos \left (\frac{s}{\sqrt{b}} \right ) \sin \left (\frac{a}{\sqrt{b}} \right ) \right ] \frac{ds}{s}\\ &= \frac{1}{\sqrt{b}} \cos \left (\frac{a}{\sqrt{b}} \right ) \int_a^\infty \sin \left (\frac{s}{\sqrt{b}} \right ) \frac{ds}{s} - \frac{1}{\sqrt{b}} \sin \left (\frac{a}{\sqrt{b}} \right ) \int_a^\infty \cos \left (\frac{s}{\sqrt{b}} \right ) \frac{ds}{s}. \end{align*} Finally, enforcing a substitution of $s \mapsto s \sqrt{b}$ in both integrals gives $$I = \frac{1}{\sqrt{b}} \cos \left (\frac{a}{\sqrt{b}} \right ) \int_{a/\sqrt{b}}^\infty \frac{\sin s}{s} \, ds - \frac{1}{\sqrt{b}} \sin \left (\frac{a}{\sqrt{b}} \right ) \int_{a/\sqrt{b}}^\infty \frac{\cos s}{s} \, ds.\tag1$$

Note that the (upper) Sine integral $\operatorname{si}(x)$ is defined as $$\operatorname{si} (x) = -\int_x^\infty \frac{\sin x}{x} \, dx,$$ and the Cosine integral $\operatorname{Ci}(x)$ is defined as $$\operatorname{Ci}(x) = - \int_x^\infty \frac{\cos x}{x} \, dx,$$ so in terms of these two functions (1) can be expressed as $$\boxed{\int_0^\infty \frac{e^{-at}}{1 + bt^2} \, dt = \frac{1}{\sqrt{b}} \sin \left (\frac{a}{\sqrt{b}} \right ) \operatorname{Ci} \left (\frac{a}{\sqrt{b}} \right ) - \frac{1}{\sqrt{b}} \cos \left (\frac{a}{\sqrt{b}} \right ) \operatorname{si} \left (\frac{a}{\sqrt{b}} \right )}$$

Answer 2

I think I have found the answer using the book Table of Integrals, Series and Products. I will post it here for more expert people to let me know if I have doe it right

$$\int^\infty_0f(r) dr = A\int^\infty_0 \frac{r}{1+Cr^4} e^{-Br^2} dr$$

If I consider $t= r^2$ and $s = B$

$$\int^\infty_0f(r) dr = \frac{A}{2}\int^\infty_0 \frac{1}{1+Ct^2} e^{-st} dt = \mathcal{L}_{f(t)} (s)$$

In other words

$$ \int^\infty_0f(r) dr = \frac{A}{2}\mathcal{L}_{f(t)} (s)$$

When $s=B$, $t = r^2$ and $f(t) = \frac{1}{1 + Ct^2}$

Then using the image below (which is given in the reference 1 below)

Considering

$$ = \frac{A C}{2}\int^\infty_0 \frac{e^{-st} }{(1/\sqrt C)^2+t^2} dt$$

$$ \int^\infty_0f(r) dr = \frac{A C}{2}\Big[ci\bigg(\frac{s}{\sqrt C}\bigg)\sin \bigg(\frac{s}{\sqrt C}\bigg) - si\bigg(\frac{s}{\sqrt C}\bigg)\cos \bigg(\frac{s}{\sqrt C}\bigg)\Big]$$

Bote however, I have one confusion that the comments says that $Re\ b > 0$ and $Re\ \mu > 0$. In my case I do not have the complex values at all so it should work fine.