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Find $$\int_0^{t} xe^{i(a x^4-bx^2)}J_0(cx)\,\mathrm dx$$

Given $a>0,\ b>0, c>0, t>0$.

I decompose it into two parts:

$$\int_0^{t} x\cos{(a x^4-bx^2)}J_0(cx)\,\mathrm dx,$$

$$\int_0^{t} x\sin{(a x^4-bx^2)}J_0(cx)\,\mathrm dx.$$

And I found some equations in Table of Integrals, Series, and Products by Gradshteyn and Ryzhik.

$$\int_0^{\infty}J_{0}(\beta x)\sin(\alpha x^2)x\,\mathrm dx=\frac{1}{2\alpha}\cos\left(\frac{\beta^2}{4\alpha}\right),$$

$$\int_0^{\infty}J_{0}(\beta x)\cos(\alpha x^2)x\,\mathrm dx=\frac{1}{2\alpha}\sin\left(\frac{\beta^2}{4\alpha}\right).$$

Now the problem is the term $x^4$ and the integration range $t$. Very appreciated if I can get help.

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  • $\begingroup$ It seems difficult to deals with the integration range but you can at least deals with the $x^4$: With the change of variables $\xi=x^2$ you have: $$\int_0^{t} x\exp{(i(a x^4-bx^2))}J_0(cx)\,\mathrm dx = \frac{1}{2}\int_0^{t^2} \exp{(i(a \xi^2-b\xi))}J_0(cx)\,\mathrm d \xi,$$ then completing the square: $$\frac{1}{2}\int_0^{t^2} \exp{(i(a \xi^2-b\xi))}J_0(cx)\,\mathrm d \xi,= \frac{e^{-i\frac{b^2}{4a}}}{2}\int_0^{t^2} \exp{\left(ia \left( \xi-\frac{b}{2} \right)^2 \right)}J_0(cx)\,\mathrm d \xi,$$ which is closer to the Gradshteyn and Ryzhik formula. $\endgroup$ – Delta-u Oct 14 '18 at 16:18

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