Proof of Laplace transform of the Bessel function

Question

I need to prove that: $\frac{1}{\sqrt{s^2+\alpha^2}} = J_0(\alpha t)$
please help

Answer 1

Well, we know that:

$$\mathcal{J}_0\left(x\right):=\frac{2}{\pi}\int_0^1\frac{\cos\left(xt\right)}{\sqrt{1-t^2}}\space\text{d}t\tag1$$

So, we also get that:

$$\mathcal{J}_0\left(\alpha t\right):=\frac{2}{\pi}\int_0^1\frac{\cos\left(\alpha tx\right)}{\sqrt{1-x^2}}\space\text{d}x\tag2$$

Now, for the Laplace transform (by using the definition):

$$\mathcal{L}_t\left[\mathcal{J}_0\left(\alpha t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty e^{-\text{s}t}\left\{\frac{2}{\pi}\int_0^1\frac{\cos\left(\alpha tx\right)}{\sqrt{1-x^2}}\space\text{d}x\right\}\space\text{d}t\tag3$$

And, so we get that:

$$\mathcal{L}_t\left[\mathcal{J}_0\left(\alpha t\right)\right]_{\left(\text{s}\right)}=\frac{2}{\pi}\int_0^\infty\int_0^1\frac{e^{-\text{s}t}\cos\left(\alpha tx\right)}{\sqrt{1-x^2}}\space\text{d}x\space\text{d}t\tag4$$

Which gives, also that (by using Fubini's theorem, changing the order of integration):

$$\mathcal{L}_t\left[\mathcal{J}_0\left(\alpha t\right)\right]_{\left(\text{s}\right)}=\frac{2}{\pi}\int_0^1\int_0^\infty\frac{e^{-\text{s}t}\cos\left(\alpha tx\right)}{\sqrt{1-x^2}}\space\text{d}t\space\text{d}x\tag5$$

So, looking for the inner integral:

$$\int_0^\infty\frac{e^{-\text{s}t}\cos\left(\alpha tx\right)}{\sqrt{1-x^2}}\space\text{d}t=\frac{1}{\sqrt{1-x^2}}\int_0^\infty e^{-\text{s}t}\cos\left(\alpha tx\right)\space\text{d}t=$$ $$\frac{1}{\sqrt{1-x^2}}\cdot\mathcal{L}_t\left[\cos\left(\alpha tx\right)\right]_{\left(\text{s}\right)}=\frac{1}{\sqrt{1-x^2}}\cdot\frac{\text{s}}{\text{s}^2+\alpha^2\cdot x^2}\tag6$$

When $\left|\Im\left(\alpha x\right)\right|<\Re\left(\text{s}\right)$

And for the outer integral:

$$\int_0^1\frac{1}{\sqrt{1-x^2}}\cdot\frac{\text{s}}{\text{s}^2+\alpha^2\cdot x^2}\space\text{d}x=\frac{\text{s}\pi\sqrt{\frac{\alpha^2}{\text{s}^2}}}{2\alpha^2\sqrt{1+\frac{\text{s}^2}{\alpha^2}}}\tag7$$

When $\Re\left(\frac{\text{s}^2}{\alpha^2}\right)\ge0\space\bigvee\space\Re\left(\frac{\text{s}^2}{\alpha^2}\right)\le-1\space\bigvee\space\frac{\text{s}^2}{\alpha^2}\not\in\mathbb{R}$

So, we get that:

$$\mathcal{L}_t\left[\mathcal{J}_0\left(\alpha t\right)\right]_{\left(\text{s}\right)}=\frac{2}{\pi}\cdot\frac{\text{s}\pi\sqrt{\frac{\alpha^2}{\text{s}^2}}}{2\alpha^2\sqrt{1+\frac{\text{s}^2}{\alpha^2}}}=\frac{\text{s}\sqrt{\frac{\alpha^2}{\text{s}^2}}}{\alpha^2\sqrt{1+\frac{\text{s}^2}{\alpha^2}}}\tag8$$

Which simplifies to $\frac{1}{\sqrt{\text{s}^2+\alpha^2}}$ when both the variables are real.

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EDIT:

$$\int_0^1\frac{1}{\sqrt{1-x^2}}\cdot\frac{\text{s}}{\text{s}^2+\alpha^2\cdot x^2}\space\text{d}x=\text{s}\int_0^1\frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\text{s}^2+\alpha^2\cdot x^2}\space\text{d}x$$

Now, substitute $x=\sin\left(\text{u}\right)$:

$$\text{s}\int_0^1\frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\text{s}^2+\alpha^2\cdot x^2}\space\text{d}x=\text{s}\int_0^\frac{\pi}{2}\frac{1}{\text{s}^2+\alpha^2\cdot\sin^2\left(\text{u}\right)}\space\text{d}\text{u}$$

Now, substitute $\text{z}=\cot\left(\text{u}\right)$:

$$\text{s}\int_0^\frac{\pi}{2}\frac{1}{\text{s}^2+\alpha^2\cdot\sin^2\left(\text{u}\right)}\space\text{d}\text{u}=-\text{s}\int_\infty^0\frac{1}{\alpha^2+\text{s}^2\cdot\left(1+\text{z}^2\right)}\space\text{d}\text{z}$$

Well, for substitute $\text{p}=\frac{\text{s}\cdot\text{z}}{\sqrt{\alpha^2+\text{s}^2}}$:

$$\int\frac{1}{\alpha^2+\text{s}^2\cdot\left(1+\text{z}^2\right)}\space\text{d}\text{z}=-\frac{1}{\text{s}\sqrt{\alpha^2+\text{s}^2}}\int\frac{1}{1+\text{p}^2}\space\text{d}\text{p}=-\frac{1}{\text{s}\sqrt{\alpha^2+\text{s}^2}}\cdot\arctan\left(\text{p}\right)+\text{C}$$

Answer 2

We start our arguments at\begin{align} J_0(\alpha x)&=\frac{1}{\pi}\int_0^\pi \cos (-\alpha x\sin \theta ) d\theta =\frac{1}{\pi}\int_0^\pi \cos (\alpha x\sin \theta ) d\theta\\ &=\frac{2}{\pi}\int_0^{\pi/2}\cos(\alpha x\sin \theta ) d\theta =\frac{2}{\pi}\int_0^{\pi/2}\cos(\alpha x\cos\varphi ) d\varphi \quad(\theta =\pi/2-\varphi ). \end{align} Then for $s>0$ \begin{align} \int_0^\infty e^{-sx}J_0(\alpha x)dx&=\frac{2}{\pi}\int_0^\infty e^{-sx}dx\int_0^{\pi/2}\cos(\alpha x\cos \varphi )d\varphi \\ &=\frac{2}{\pi}\int_0^{\pi/2}d\varphi \int_0^\infty e^{-sx}\cos(\alpha x\cos \varphi )dx\\ &=\frac{2}{\pi}\int_0^{\pi/2} \frac{s}{\alpha ^2\cos^2\varphi +s^2}d\varphi\\ &=\frac{1}{\sqrt{\alpha ^2+s^2}} .\tag{1} \end{align}

Here we used the well known results $$\int_0^\infty e^{-sx}\cos( px)dx=\frac{s}{p^2+s^2}\quad (s>0)$$ and $$ \int_0^{\pi/2} \frac{d\theta }{q^2+\cos^2\theta }=\frac{\pi}{2q\sqrt{1+q^2}}\quad(q>0).$$ From $(1)$ we have $$ \mathscr{L}^{-1}\left(t\,;\frac{1}{\sqrt{s^2+\alpha ^2}}\right)=J_0(\alpha t).$$

Answer 3

It's not bad to know this series of $J_0(x)$ sometimes is useful, \begin{eqnarray*} {\cal L}(y)&=&\frac{1}{\sqrt{s^2+1}}\\ &=&\frac{1}{s}(1+\frac{1}{s^2})^{-\frac12}~~~~~~~~~~~~~~~~\text{Binomial}\\ &=&\frac{1}{s}\Big(1-\frac12(\frac{1}{s^2})-\frac{1}{2!}\frac12\frac32(\frac{1}{s^2})^2+\frac{1}{3!}\frac12\frac32\frac52(\frac{1}{s^2})^3+\cdots\Big) \\ &=&\sum_{n=0}^\infty\frac{(-1)^n(2n)!}{2^{2n}(n!)^2s^{2n+1}}\\ y&=&\sum_{n=0}^\infty\frac{(-1)^n}{2^{2n}(n!)^2}x^{2n}\\ &=&1-\frac{x^2}{2^2}+\frac{x^4}{2^24^2}-\frac{x^6}{2^24^26^2}+\cdots\\ &=& J_0(x) \end{eqnarray*}