Calculate a definite integral involving sin and exp I got these definite integrals from the moments.
I am required to calculate the definite integrals $$\int_{0}^{\infty}x^kf(x)dx$$, where $f(x)=e^{-x^{\frac{1}{4}}}sin(x^{\frac{1}{4}})$ and $k\in\mathbb N$.
I have been through a tough time on trying to calculate it. I tried to change variable and integrate by parts, but it seems not working.
Any help will be appreciated.
Edit:
I first used the change of variable turing the definite integral into 
$$4\int_0^{\infty}y^{4k+3}e^{-y}sin(y)dy$$, where $y=x^{\frac{1}{4}}.$
And then when I tried to integrate by parts, two new terms will come out, which are $y^{4k+2}sin(y)$ and $y^{4k+3}cos(y).$ I noticed that we can never eliminate the power of $y$ from the second term, so I had no idea how to go on.
 A: Hint The appearance of the quantity $\require{cancel}x^{1 / 4}$ inside the arguments of $\exp$ and $\sin$ suggest the substitution $$x = u^4, \qquad dx = 4 u^3 du,$$ which transforms the integral into
$$4 \int_0^\infty u^{4 k + 3} e^{-u} \sin u \,du
= 4 \int_0^\infty u^{4 k + 3} e^{-u} \operatorname{Im}(e^{i u}) \,du
= 4 \operatorname{Im} \int_0^{\infty} u^{4 u + 3} e^{(-1 + i) u} du .$$
The form of the integrand suggests applying integration by parts. Doing so for
$$\int_0^{\infty} u^m e^{(-1 + i) u} du$$ with $p = u^m$, $dq = e^{(-1 + i) u} du$ gives
$$\cancelto{0}{\left.u^m \cdot \frac{1}{-1 + i} e^{(-1 + i) u} \right\vert_0^\infty} - \int_0^\infty m u^{m - 1} \frac{1}{-1 + i} e^{(-1 + i) u} du .$$
So, if we denote $I_m := \int_0^{\infty} u^m e^{(-1 + i) u} du ,$ the integrals $I_m$ satisfy the reduction formula
$$I_m = -\frac{m e^{-3 \pi i / 4}}{\sqrt{2}} I_{m - 1}.$$
A: An alternative approach. Here your integral is:
\begin{equation}
 I = 4\int_0^\infty x^{4k + 3} e^{-x} \sin(x)\:dx\nonumber
\end{equation}
Here we will employ Feynman's Trick by introducing the function 
\begin{equation}
 J(t) = 4\int_0^\infty x^{4k + 3} e^{-tx} \sin(x)\:dx\nonumber
\end{equation}
We observe that $J(1) = I$. Now:
\begin{equation}
 \frac{\partial}{\partial t} e^{-tx} = -x e^{-tx} \Longrightarrow \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} e^{-tx} = \left(-1\right)^{4k + 3} x^{4k + 3}e^{-tx} = -x^{4k + 3}e^{-tx}
\end{equation}
And thus, 
\begin{equation}
 J(t) = 4\int_0^\infty x^{4k + 3} e^{-tx} \sin(x)\:dx =  4\int_0^\infty  -\frac{\partial^{4k + 3}}{\partial t^{4k + 3}} e^{-tx} \sin(x)\:dx = -4\int_0^\infty  \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} e^{-tx} \sin(x)\:dx\nonumber
\end{equation}
By Leibniz's Integral Rule:
\begin{equation}
 J(t) =-4\int_0^\infty  \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} e^{-tx} \sin(x)\:dx =  -4\frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \int_0^\infty  e^{-tx}\sin(x)\:dx \nonumber
\end{equation}
Now 
\begin{equation}
\int e^{ax}\sin\left(bx\right)\:dx = \frac{e^{ax}}{a^2 + b^2}\left(a\sin(bx) - b\cos(bx) \right) + C\nonumber
\end{equation}
Thus our integral becomes 
\begin{align}
 J(t) &=-4 \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \int_0^\infty  e^{-tx}\sin(x)\:dx = -4\frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \left[\frac{e^{-tx}}{t^2 + 1}\left(-t\sin(x) - \cos(x) \right)  \right]_0^{\infty}\nonumber \\
&= -4 \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \left[\frac{1}{t^2 + 1} \right] \nonumber
\end{align}
And so we can express the solution to $I$ as 
\begin{equation}
I = J(1) = -4\frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \frac{1}{t^2 + 1}\bigg|_{t = 1}
\end{equation}
