# 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.

• ops the range should be $[0,\infty]$ – Sam Wong Mar 12 at 6:45
• A better choice for integration by parts is $u = y^{4 k + 3}$, $dv = e^{-y} \sin y \,dy$, but even this becomes a little messy, owing to the fact that $k$ is arbitrary. It simplifies things some to write $\sin y$ in terms of complex numbers, so that problem amounts to integrating $y^{4 k + 3} e^{a y}$ for some constant $a$. Then, the integrating step in integration by parts becomes a little more tractable. See my answer. – Travis Mar 12 at 7:17

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 $$I_m = -\frac{m e^{-3 \pi i / 4}}{\sqrt{2}} I_{m - 1}.$$
• Thanks! But how did you come up an idea that turns sin $u$ into Im($e^{iu}$)? This step is so amazing. – Sam Wong Mar 12 at 7:19
• You're welcome. This is a standard trick for integrals that involve both $\exp$ and $\sin$ or $\cos$---even when $\exp$ doesn't appear in the integrand, rewriting $\sin$ or $\cos$ in terms of the complex exponential sometimes simplifies things. – Travis Mar 12 at 7:21
• @SamWong FYI, I recall listening to a talk from some of the developers involved with writing the Maple symbolic & numeric computing environment fairly shortly after it was initially created (this would have been about $1985$). They described some of what it did behind the scenes, including tricks like what Travis used to convert parts of certain integrals to the real or imaginary part of a complex function which is easier to integrate. After all these years, Maple has likely changed quite a bit in terms of how it works internally, but I suspect certain "tricks" like this are still being used. – John Omielan Mar 12 at 7:34
An alternative approach. Here your integral is: $$$$I = 4\int_0^\infty x^{4k + 3} e^{-x} \sin(x)\:dx\nonumber$$$$ Here we will employ Feynman's Trick by introducing the function $$$$J(t) = 4\int_0^\infty x^{4k + 3} e^{-tx} \sin(x)\:dx\nonumber$$$$ We observe that $$J(1) = I$$. Now: $$$$\frac{\partial}{\partial t} e^{-tx} = -x e^{-tx} \Longrightarrow \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} e^{-tx} = \left(-1\right)^{4k + 3} x^{4k + 3}e^{-tx} = -x^{4k + 3}e^{-tx}$$$$ And thus, $$$$J(t) = 4\int_0^\infty x^{4k + 3} e^{-tx} \sin(x)\:dx = 4\int_0^\infty -\frac{\partial^{4k + 3}}{\partial t^{4k + 3}} e^{-tx} \sin(x)\:dx = -4\int_0^\infty \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} e^{-tx} \sin(x)\:dx\nonumber$$$$ By Leibniz's Integral Rule: $$$$J(t) =-4\int_0^\infty \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} e^{-tx} \sin(x)\:dx = -4\frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \int_0^\infty e^{-tx}\sin(x)\:dx \nonumber$$$$ Now $$$$\int e^{ax}\sin\left(bx\right)\:dx = \frac{e^{ax}}{a^2 + b^2}\left(a\sin(bx) - b\cos(bx) \right) + C\nonumber$$$$ Thus our integral becomes \begin{align} J(t) &=-4 \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \int_0^\infty e^{-tx}\sin(x)\:dx = -4\frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \left[\frac{e^{-tx}}{t^2 + 1}\left(-t\sin(x) - \cos(x) \right) \right]_0^{\infty}\nonumber \\ &= -4 \frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \left[\frac{1}{t^2 + 1} \right] \nonumber \end{align} And so we can express the solution to $$I$$ as $$$$I = J(1) = -4\frac{\partial^{4k + 3}}{\partial t^{4k + 3}} \frac{1}{t^2 + 1}\bigg|_{t = 1}$$$$