Laplace Transform of integrals

Question

How do I prove that the Bessel function $J_0 = {1\over \pi}\int_0^\pi \cos(x\cos(t))dt $ using Laplace transforms? I was able to solve a similar problem by finding the transform of the integrand by regarding the $t$ terms as constants and then finding the inverse, but I just don't seem to understand the logic. Any help appreciated.

Answer 1

\begin{align} {\cal L}(J_0) &= \int_0^\infty e^{-sx}{1\over \pi}\int_0^\pi \cos(x\cos(t))dt \, dx \\ &= \int_0^\infty e^{-sx}{1\over \pi}\int_0^\pi \dfrac12\left(e^{ix\cos t}+e^{-ix\cos t}\right)dt \, dx \\ &= \dfrac{1}{2\pi} \int_0^\infty\int_0^\pi \left(e^{ix\cos t-sx}+e^{-ix\cos t-sx}\right)dt \, dx \\ &= \dfrac{1}{2\pi} \int_0^\pi \int_0^\infty \left(e^{ix\cos t-sx}+e^{-ix\cos t-sx}\right)dx \, dt \\ &= \dfrac{1}{2\pi} \int_0^\pi \dfrac{e^{ix\cos t-sx}}{i\cos t-s}+\dfrac{e^{-ix\cos t-sx}}{-i\cos t-s}\Big|_0^\infty \, dt \\ &= \dfrac{1}{2\pi} \int_0^\pi \dfrac{1}{s-i\cos t}+\dfrac{1}{-s-i\cos t} \, dt \\ &= \dfrac{s}{\pi} \int_0^\pi \dfrac{1}{s^2+\cos^2t} \, dt \\ &= \dfrac{s}{\pi} \dfrac{\pi}{s\sqrt{s^2+1}}\\ &= \dfrac{1}{\sqrt{s^2+1}} \end{align}

Answer 2

In general:

For the Laplace transform:

• $$\text{F}\left(\text{s}\right):=\mathscr{L}_x\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{f}\left(x\right)\cdot e^{-\text{s}x}\space\text{d}x\tag1$$

For the Bessel function:

• $$\mathcal{J}_\alpha\left(x\right):=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\left(\frac{x}{2}\right)^{\alpha+2\text{n}}\tag2$$

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So, we get:

$$\text{F}_\alpha\left(\text{s}\right)=\mathscr{L}_x\left[\mathcal{J}_\alpha\left(x\right)\right]_{\left(\text{s}\right)}=\int_0^\infty\mathcal{J}_\alpha\left(x\right)\cdot e^{-\text{s}x}\space\text{d}x=$$ $$\int_0^\infty\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\left(\frac{x}{2}\right)^{\alpha+2\text{n}}\cdot e^{-\text{s}x}\space\text{d}x=$$ $$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\int_0^\infty\left(\frac{x}{2}\right)^{\alpha+2\text{n}}\cdot e^{-\text{s}x}\space\text{d}x=$$ $$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\int_0^\infty\frac{x^{\alpha+2\text{n}}}{2^{\alpha+2\text{n}}}\cdot e^{-\text{s}x}\space\text{d}x=$$ $$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{1}{2^{\alpha+2\text{n}}}\cdot\int_0^\infty x^{\alpha+2\text{n}}\cdot e^{-\text{s}x}\space\text{d}x\tag3$$

Now, using the qth power (for complex q) Laplace transform:

$$\int_0^\infty x^{\alpha+2\text{n}}\cdot e^{-\text{s}x}\space\text{d}x=\frac{\Gamma\left(1+\alpha+2\text{n}\right)}{\text{s}^{1+\alpha+2\text{n}}}\tag4$$

When $\Re\left(\text{s}\right)>0\space\wedge\space\Re\left(\alpha+2\text{n}\right)>-1$

So, we end up with:

$$\text{F}_\alpha\left(\text{s}\right)=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\Gamma\left(1+\alpha+2\text{n}\right)}{\text{s}^{1+\alpha+2\text{n}}}\tag5$$

Now, when $\alpha=0$:

$$\text{F}_0\left(\text{s}\right)=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\text{n}\right)}\cdot\frac{1}{2^{2\text{n}}}\cdot\frac{\Gamma\left(1+2\text{n}\right)}{\text{s}^{1+2\text{n}}}=\frac{1}{\text{s}}\cdot\frac{1}{\sqrt{1+\frac{1}{\text{s}^2}}}\tag6$$

When $\frac{1}{\left|\text{s}\right|}<1\space\wedge\space\Re\left(\text{s}\right)>0$