Closed form of $ x^3\sum_{k=0}^{\infty} \frac{(-x^4)^k}{(4k+3)(1+2k)!} $

Question

I was trying to find a closed form of $$ f(x) = x^3\sum_{k=0}^{\infty} \frac{(-x^4)^k}{(4k+3)(2k+1)!} $$

$f(x)$ converges for all $x$ by the ratio test.

I began by making the it look like a power series and then differentiating $$ f(x) = \sum_{k=0}^{\infty} \frac{(-1)^kx^{4k+3}}{(4k+3)(2k+1)!} $$

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \sum_{k=0}^{\infty} \frac{(-1)^kx^{4k+3}}{(4k+3)(2k+1)!} $$

$$ \frac{\partial f}{\partial x} = \sum_{k=0}^{\infty} \frac{(-1)^kx^{4k+2}}{(2k+1)!} = \sum_{k=0}^{\infty} \frac{(-1)^k(x^2)^{2k+1}}{(2k+1)!} $$

$$ \frac{\partial f}{\partial x} = \sin(x^2) $$

$$f(x) = c + \int \sin x^2 \mathrm{d}x $$

Is this correct? Can it be solved simpler than this? Can this result be simplified further? I'm not sure how to take the integral of $\sin x^2 $.

Answer 1

Initial condition of $f(0)=0$ gives $c=0$. The rest may be done with the complex definition of sine and the error function:

$$f(x)=\int_0^x\sin(t^2)\ dt=\int_0^x\frac{e^{it^2}-e^{-it^2}}{2i}\ dt=\frac{\sqrt\pi}4\left(e^{-5\pi i/8}\operatorname{erf}(e^{3\pi i/8}x)+e^{3\pi i/8}\operatorname{erf}(e^{\pi i/8}x)\right)$$

which is honestly a horrible expression.

An alternative expression:

$$f(x)=\Im\int_0^xe^{it^2}\ dt=\Im\left(\frac{\sqrt\pi}2e^{\pi i/8}\operatorname{erf}(e^{-\pi i/8}x)\right)$$

Answer 2

$\sin x^2$ is one of those functions whose primitive can not be expressed in terms of elementary functions. Nevertheless, your idea is correct. You can improve your answer by writing

$$f(x) = f(0) + \int_0^x \sin t^2 dt = \int_0^x \sin t^2 dt,$$

as $f(0) =0$.