How to prove $\int_{0}^{\infty} \sqrt{x} J_{0}(x)dx = \sqrt{2} \frac{\Gamma(3/4)}{\Gamma(1/4)}$

Question

I was reading a paper in which they used the following relation and I am curious to know how it can be proved.

$$\int_{0}^{\infty} \sqrt{x} J_{0}(x)dx = \sqrt{2} \frac{\Gamma(3/4)}{\Gamma(1/4)}$$

where $J_{0}(x)$ is a Bessel function of the first kind and $\Gamma(x)$ is the gamma function.

I tired to prove it using the integral form of the Bessel function as: $J_{0}(r)=\int_{0}^{2\pi}e^{-ir \cos(\theta)}d\theta $ and the integral form of Gamma function as: $\Gamma(r)=\int_{0}^{\infty}x^{r-1}e^{-x}dx$ but I couldn't. Then, I tried to prove it using the series representation of the Bessel function as $J_{0}(x)= \sum_{k=0}^{\infty} \frac{(-1)^k}{(\Gamma(k+1))^2} (\frac{x}{2})^{2k}$ in the integral but I couldn't still prove it.

I don't know if a complex contour integral formula for the Bessel function and Gamma function can be useful to prove this, but anyway I'm trying every thing.

Could you please help me to prove this relation?

Thank you

Answer 1

Well, we have:

$$\mathcal{I}_{\space\frac{1}{2}}:=\int\limits_0^\infty x^\frac{1}{2}\cdot\mathscr{J}_0\left(x\right)\space\text{d}x\tag1$$

Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:

$$\mathcal{I}_{\space\frac{1}{2}}=\int\limits_0^\infty\mathscr{L}_x\left[\mathscr{J}_0\left(x\right)\right]_{\left(\text{s}\right)}\cdot\mathscr{L}_x^{-1}\left[x^\frac{1}{2}\right]_{\left(\text{s}\right)}\space\text{d}\text{s}\tag2$$

Using table of selected Laplace transforms:

$$\mathcal{I}_{\space\frac{1}{2}}=\int\limits_0^\infty\frac{1}{\sqrt{1+\text{s}^2}}\cdot\frac{1}{\text{s}^{1+\frac{1}{2}}}\cdot\frac{1}{\Gamma\left(-\frac{1}{2}\right)}\space\text{d}\text{s}=-\frac{1}{2\sqrt{\pi}}\int\limits_0^\infty\frac{1}{\sqrt{1+\text{s}^2}}\cdot\frac{1}{\text{s}^\frac{3}{2}}\space\text{d}\text{s}\tag3$$

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And now, we can use:

$$\int\limits_0^\infty\frac{1}{x^\text{n}}\cdot\frac{1}{\sqrt{1+x^2}}\space\text{d}x=\frac{\displaystyle\Gamma\left(\frac{1}{2}-\frac{\text{n}}{2}\right)\cdot\Gamma\left(\frac{\text{n}}{2}\right)}{\displaystyle2\sqrt{\pi}}\tag4$$

When $0<\Re\left(\text{n}\right)<1$.

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So, we end up with:

$$\mathcal{I}_{\space\frac{1}{2}}=-\frac{\displaystyle1}{\displaystyle2\sqrt{\pi}}\cdot\frac{\displaystyle\Gamma\left(\frac{1}{2}-\frac{3}{2}\cdot\frac{1}{2}\right)\cdot\Gamma\left(\frac{3}{2}\cdot\frac{1}{2}\right)}{\displaystyle2\sqrt{\pi}}=\frac{\displaystyle\Gamma^2\left(\frac{3}{4}\right)}{\displaystyle\pi}\tag5$$

Answer 2

The Bessel function of the first kind has a series expansion, $$J_0(x)=\sum_{k=0}^\infty\frac{(-1)^k}{(k!)^2}\left(\frac{x}{2}\right)^{2k}$$ hence by the slight extension of Ramanujan's master theorem, $$\int_0^\infty x^{s-1}J_0(x)\ dx=2^{s-1}\frac{\Gamma\left(\dfrac{s}{2}\right)}{\Gamma\left(1-\dfrac{s}{2}\right)}$$ taking $s=3/2$, $$\int_0^\infty x^{1/2}J_0(x)\ dx=2^{1/2}\frac{\Gamma\left(\dfrac{3}{4}\right)}{\Gamma\left(\dfrac{1}{4}\right)}.$$