Integral of $\sin (x^3)dx$
$$\int \sin (x^3)dx$$ I have tried some substitutions, but I haven't reached the goal... Can you help me?
$\endgroup$ 35 Answers
$\begingroup$By combining Euler's formula with the integral expression for the $\Gamma$ function, we have, for $n>1$ $$\int_0^\infty\sin(x^n)~dx~=~\bigg(\frac1n\bigg){\large!}~\sin\frac\pi{2n}$$ and $$\int_0^\infty\cos(x^n)~dx~=~\bigg(\frac1n\bigg){\large!}~\cos\frac\pi{2n}$$ Given the fact that your integral is indefinite, its expression is very similar, but it involves the incomplete $\Gamma$ function, rather than the classical one.
$\endgroup$ 1 $\begingroup$Possibly this is not useful to you, but...
$$\sin(x^3)=\sum_{n=0}^\infty(-1)^n\frac{x^{6n+3}}{(2n+1)!}$$
so
$$\int\sin(x^3)dx=C+\sum_{n=0}^\infty(-1)^n\frac{x^{6n+4}}{(2n+1)!(6n+4)}$$
$\endgroup$ $\begingroup$Your integral is the imaginary part of the Fresnel integral:
$$ \int \exp{(\mathrm{i} x^3)} \, \mathrm{d}x = x \, {}_1F_1(1/3,4/3,\mathrm{i}x^3), $$ where I have substituted $n=3$ and $m = 0$ from here·
$\endgroup$ $\begingroup$First note that
$$\int sin(x^3)dx = \frac{1}{2}\left(\int i e^{ix^3}dx - \int i e^{-ix^3}dx\right)$$
For the first integral
$$I_1 = \int ie^{ix^3}dx$$
change variables with $u=ix^3$, $x = -\sqrt[3]{iu}$ and
$$dx = -\frac{1}{3i} \frac{du}{\sqrt[3]{u^2}}$$
then
$$I_1 = -\frac{1}{3} \int u^{-2/3} e^u du$$
Now with the incomplete gamma function
$$\gamma(s,z) = \int_{0}^{z} t^{s-1} e^{-t} dt$$
we can write $I_1$ as
$$I_1 = -\frac{1}{3} \frac{(-u)^{2/3}}{u^{2/3}}\gamma(1/3, -u)$$
and substituting back $u=ix^3$
$$I_1 = \frac{(-ix^3)^{2/3}}{3x^2} \gamma(1/3, -ix^3)$$
Similarly for the second integral
$$I_2 = \int i e^{-ix^3}dx$$
we have
$$I_2 = - \frac{(ix^3)^{2/3}}{3x^2} \gamma(1/3, ix^3)$$
Finally
$$\int sin(x^3)dx = \frac{(-ix^3)^{2/3} \gamma(1/3, -ix^3) + (ix^3)^{2/3} \gamma(1/3, ix^3)}{6x^2}+c$$
$\endgroup$ $\begingroup$How to integrate:
$$\int \sin (x^3)dx$$
Start with a series representation for:
$$f(x) = sin(x)$$
The taylor's series representation of sin(x):
$$ f(x) = sin(x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} $$
Substitute $x \rightarrow x^3$$$ f(x^3) = sin(x^3) = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} (x^3)^{2n+1} $$
Integrate:
$$ I = \int f(x^3) dx = \int sin(x^3) dx = \int \bigg( \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} (x^3)^{2n+1} \bigg) dx $$
$$ I= \sum^{\infty}_{n=0} \Bigg(\frac{(-1)^n}{(2n+1)!} \bigg( \int (x^3)^{2n+1} dx \bigg)\Bigg) $$
$$ I = \sum^{\infty}_{n=0} \Bigg(\frac{(-1)^n}{(2n+1)!} \bigg( \int (x^3)^{2n+1} dx \bigg)\Bigg) $$
$$ I = \sum^{\infty}_{n=0} \Bigg(\frac{(-1)^n}{(2n+1)!} \bigg( \int x^{6n+3} dx \bigg)\Bigg) $$
$$I = \sum_{n=0}^\infty(-1)^n\frac{x^{6n+4}}{(2n+1)!(6n+4)} + C$$
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