Integral $\int x\sin(y^4) \operatorname dy$

Question

How integrate $\int x \sin(y^4) \operatorname dy$?

There's something called gamma appears when you integrate but I can't seem to understand what is the function gamma means.

Answer 1

Well, we have:

$$\mathscr{I}_{\space\text{n}}\left(x,\text{y}\right):=\int x\sin\left(\text{y}^\text{n}\right)\space\text{d}\text{y}=x\int\sin\left(\text{y}^\text{n}\right)\space\text{d}\text{y}\tag1$$

Using:

$$\sin\left(\text{y}^\text{n}\right)=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\left(1+2\text{k}\right)!}\cdot\left(\text{y}^\text{n}\right)^{1+2\text{k}}=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\left(1+2\text{k}\right)!}\cdot\text{y}^{\text{n}\cdot\left(1+2\text{k}\right)}\tag2$$

So, we get:

$$\mathscr{I}_{\space\text{n}}\left(x,\text{y}\right)=x\int\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\left(1+2\text{k}\right)!}\cdot\text{y}^{\text{n}\cdot\left(1+2\text{k}\right)}\space\text{d}\text{y}=x\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\left(1+2\text{k}\right)!}\int\text{y}^{\text{n}\cdot\left(1+2\text{k}\right)}\space\text{d}\text{y}\tag3$$

And, we know that:

$$\int\text{a}^\text{b}\space\text{d}\text{a}=\frac{\text{a}^{1+\text{b}}}{1+\text{b}}+\text{C}\tag4$$

So, we get:

$$\mathscr{I}_{\space\text{n}}\left(x,\text{y}\right)=x\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\left(1+2\text{k}\right)!}\cdot\frac{\text{y}^{1+\text{n}\cdot\left(1+2\text{k}\right)}}{1+\text{n}\cdot\left(1+2\text{k}\right)}+\text{C}\tag5$$

Answer 2

First of all, your solution might not be able to be written as basic elementary functions. Now, we know that $\sin(y^4)$ is bounded and continuous. Hence, it is bounded on any bounded interval of $\mathbb{R}$.

Now, observe that on a closed interval $[-x,x]$ for any $x \in \mathbb{R}$, given $\varepsilon>0$, $f(y)=\sin(y^4)$ has $P_{n}(y)$ a polynomial degree of $n$ such that $|f(y)-P_{n}(y)|<\varepsilon$ according Weierstrass Approximation Theorem.

Now, we know that $f(y)=\sum_{i=0}^{\infty}\frac{f^{(i)}(0)}{i!}y^{i}$. However, since we bound $f(y)$ on a closed interval $[-x,x]$, we know that we can interchange the limit and integral operator. Hence, given any $t \in (0,x)$, we know have
$$\int_{0}^{t}f(y)dy = \int_{0}^{t}sin(y^4)dy$$ $$\int_{0}^{t}f(y)dy = \int_{0}^{t} \lim\limits_{n\to\infty}P_{n}(y)dy$$ $$\int_{0}^{t}f(y)dy = \lim\limits_{n\to\infty}\int_{0}^{t} P_{n}(y)dy$$ $$\int_{0}^{t}f(y)dy = \lim\limits_{n\to\infty}\int_{0}^{t} \sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}y^{i}dy$$ $$\int_{0}^{t}f(y)dy = \lim\limits_{n\to\infty}\sum_{i=0}^{n}\int_{0}^{t} \frac{f^{(i)}(0)}{i!}y^{i}dy$$ $$\int_{0}^{t}f(y)dy = \lim\limits_{n\to\infty}\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}\int_{0}^{t} y^{i}dy$$ $$\int_{0}^{t}f(y)dy = \lim\limits_{n\to\infty}\sum_{i=0}^{n}[\frac{f^{(i)}(0)}{i!}\frac{1}{i+1}t^{i+1}]$$ $$\int_{0}^{t}f(y)dy = \lim\limits_{n\to\infty}\sum_{i=0}^{n}[\frac{f^{(i)}(0)}{(i+1)!}t^{i+1}]$$ Thus, we can conclude that $\int \sin(t^4)dt = \sum_{i=0}^{\infty}\frac{f^{(i)}(0)}{\Gamma(i+2)}t^{i+1}+C$ for some constant C in which $\Gamma(\alpha+1)=\alpha!$