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I was doing a similar looking integral and wanted to ask what will be the general way doing it. I don’t know how to go about doing it, can anyone please help? $$f_{n_1n_2n_3}(\alpha, \beta)=\int_0^{\infty}\mathrm{e}^{-x^{n_1}}\sin(\alpha x^{n_2})\cos(\beta x^{n_3})\,dx$$ $n_1, n_2, n_3, \alpha, \beta \in \mathbb{Z}$

I tried to use the complex number definition of $\sin(x)$ and $\cos(x)$ but then got stuck with the following: $$\frac{1}{4}\left[\int_0^{\infty}\mathrm{e}^{i\alpha x^{n_2}+i\beta x^{n_3}-x^{n_1}}\,dx+\int_0^{\infty}\mathrm{e}^{i\alpha x^{n_2}-i\beta x^{n_3}-x^{n_1}}\,dx-\int_0^{\infty}\mathrm{e}^{-i\alpha x^{n_2}+i\beta x^{n_3}-x^{n_1}}\,dx-\int_0^{\infty}\mathrm{e}^{-i\alpha x^{n_2}-i\beta x^{n_3}-x^{n_1}}\,dx\right]$$

Asv
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    Are $n_1$, $n_2$, $n_3$, $\alpha$, and $\beta$ integers? – Ty. Jun 05 '20 at 12:54
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    Likely no chance: Mathematica fails even with $$\int_0^\infty e^{-x^3}\cos(2x^3)\sin(3x^2),dx$$. – user64494 Jun 07 '20 at 05:25
  • @user64494 I gave that input to wolfram alpha and it gave me the output 0.320266, I don’t know but I think that with some luck it would be solvable using generalised fresnel integrals – Asv Jun 07 '20 at 11:20
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    I agree with @user64494 ... there is no reason to think $$\int_0^\infty e^{-x^3}\cos(2x^3)\sin(3x^2),dx$$ has a closed form. – GEdgar Jun 08 '20 at 13:38

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The present question generalizes this one to trigonometric functions of an arbitrary monomial argument. There may be no general closed-form expression. If the exponents $n_k$ are non-negative, then the integrand isn't singular. Some particular cases involving polynomial exponentials and trigonometric functions can be found here and there. If $n_1$, $n_2$, $n_3$ belong to $\lbrace 0,1,2\rbrace$, then some other converging integrals may be expressed analytically. Indeed, if the integrand is invariant under $x \mapsto -x$ (i.e., $n_1$, $n_2$ are even integers), then we may rewrite the integral as $$ f_{n_1n_2n_3}(\alpha, \beta) = \frac12\int_{-\infty}^\infty \exp(-x^{n_1})\sin(\alpha x^{n_2})\cos(\beta x^{n_3}) \, \text d x \, . $$ Using the trigonometric identity $$ \sin(\alpha x^{n_2})\cos(\beta x^{n_3}) = \tfrac12 \big(\sin(\alpha x^{n_2}-\beta x^{n_3}) + \sin(\alpha x^{n_2}+\beta x^{n_3})\big) $$ along with Euler's formula, the previous integral is rewritten as a linear combination of integrals of the form $$ I_{a,b} = \int_{-\infty}^\infty \exp\!\big({-x}^{n_1} + \text i\alpha(-1)^a x^{n_2} + \text i \beta(-1)^b x^{n_3}\big) \, \text d x $$ where $a$, $b$ belong to $\lbrace 0,1\rbrace$. Generalized Gaussian integrals from quantum field theory provide analytic expressions for particular values of $n_1$, $n_2$ in $\lbrace 0,2\rbrace$ and $n_3$ in $\lbrace 0,1,2\rbrace$.

EditPiAf
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