$\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0$

Question

I'd like to show that for all positive integers $n$ we have $$I\left( n \right)=\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0.$$

This is true after some computer experiments, besides after setting $x={{u}^{4}}$ we get

$$I\left( n \right)=4\int_{0}^{\infty }{{{u}^{4n+3}}\sin \left( u \right)\exp \left( -u \right)du}.$$

But how to proceed? Integration by parts maybe?

Answer 1

A Laplace transform is easier than integrating by parts. Consider the function $$ F_n(s) = \int_{0}^\infty t^{4n+3}e^{-st}dt = \frac{(4n+3)!}{s^{4n+4}}. $$ Then $I(n) = \mathrm{Im}[F_n(1-i)]$. Since $$ F_n(1-i) = \frac{(4n+3)!}{(1-i)^{4n+4}} = \frac{(4n+3)!}{(-4)^{n+1}} \in \mathbb R, $$ $I(n) = 0$.

Answer 2

Using contour integration, with the contour being a 'pie slice', composed of lines $$z_1(t) = t,\qquad t\in[0,R]$$ $$z_2(t) = t(1-i),\qquad t\in[0,\frac{R}{\sqrt{2}}]$$ $$z_3(t) = Re^{it},\qquad t\in[-\frac\pi 4,0] $$ and consideringing $$ \oint_C z^{4n+3} e^{-z} dz = 0$$ you can prove that $$ (1-i)^{4n+4}\int_0^\infty u^{4n+3} e^{-u(1-i)} du = \int_0^\infty u^{4n+3} e^{-u} du $$ so $$ \int_0^\infty u^{4n+3} e^{-u(1-i)} du = \frac{1}{(-4)^{n+1}} \int_0^\infty u^{4n+3} e^{-u} du = \frac{(4n+3)!}{(-4)^{n+1}} \in \mathbb R$$ Therefore $$I(n) = {\rm Im} \Big(\int_0^\infty 4u^{4n+3} e^{iu} e^{-u} du \Big) = {\rm Im} \Big(4\int_0^\infty u^{4n+3} e^{-u(1-i)} du \Big) = 0$$

Answer 3

First define for any integer $n$

$$ {{A}_{n}}=\int_{0}^{\infty }{{{x}^{n}}{{e}^{-x}}\sin \left( x \right)dx}\quad and\quad {{B}_{n}}=\int_{0}^{\infty }{{{x}^{n}}{{e}^{-x}}\cos \left( x \right)dx} $$ Now the substitution $u={{x}^{1/4}}$ reduces the integral in question to $4{{A}_{4n+3}}$.

Using integration by parts any one can verify the recurrence relation: $$ \left\{ \begin{align} & {{A}_{n}}=\frac{n}{2}\left( {{A}_{n-1}}+{{B}_{n-1}} \right) \\ & {{B}_{n}}=\frac{n}{2}\left( {{B}_{n-1}}-{{A}_{n-1}} \right) \\ \end{align} \right. $$ Solving this with initial conditions: ${{A}_{0}}=1/2\quad and\quad {{B}_{0}}=1/2$ you get ${{A}_{n}}=0$ for any integer $n$ such that $n\equiv 3\left( \bmod 4 \right)$ which means that $4{{A}_{4n+3}}=0$.

Answer 4

The following will also evaluate the integral for real values of $n$:

With the substitution you mentioned, we have $$I\left( n \right)=4\int_{0}^{\infty }{{{u}^{4n+3}}\sin \left( u \right)\exp \left( -u \right)du} = 4 \Im \left\{ \int_{0}^{\infty }u^{4n+3}\exp \left( u(i-1) \right)du \right\}=4 \Im \{J(n)\}$$ With $s=4(n+1)$ and the series representation of the exponential, we can write $$J(n) = \int_0^\infty dx~ x^{s-1} \sum_{k=0}^\infty (i-1)^k \frac{x^k}{k!}=\int_0^\infty dx~ x^{s-1} \sum_{k=0}^\infty (1-i)^k \frac{(-x)^k}{k!}=\int_0^\infty dx~ x^{s-1} f(x)$$ Now, let's use Ramanujan's master theorem! It tells us that $$J(n)=\Gamma(s)\varphi(-s)$$ where $\Gamma(s)$ is the Gamma function and $\varphi(k)=(1-i)^k$. Putting back $s=4(n+1)$ we obtain $$J(n)=\Gamma(4n+4) (1-i)^{-4n-4}\\=\Gamma(4n+4) \frac{1}{(\sqrt{2})^{4n+4}} \left( \frac{1-i}{\sqrt{2}}\right)^{-4n-4}\\=\Gamma(4n+4) \frac{1}{4^{n+1}}\left(\exp(-i\pi/4) \right)^{-4n-4} \\ = \Gamma(4n+4) \frac{1}{4^{n+1}} e^{i\pi(n+1)}$$ Hence, finally $$I(n) = 4 \Im \{J(n)\} = 4\Gamma(4n+4) \frac{1}{4^{n+1}} \sin(\pi(n+1))$$ which is $$I(n)=-\frac{\Gamma(4n+4)}{4^n}\sin(n\pi)$$ which, for integers $n$ yields $$I(n)=0$$ but the result should also hold for real values (which I checked for n=1.4 numerically).

Answer 5

Let $$I_n = \int_0^\infty x^n \sin (\sqrt[4]{x}) \exp (\sqrt[4]{x}) \, dx, \qquad n \in \mathbb{N}.$$ After enforcing a substitution of $x \mapsto \sqrt[4]{x}$ one has $$I_n = 4 \int_0^\infty x^{4n + 3} e^{-x} \sin x \, dx.$$

The following useful property for the Laplace transform will now be employed to evalaute the integral: $$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$ Noting that $$\mathcal{L} \{\ x^{4n + 3} \sin x \}(t) = \frac{(4n + 3)!}{(1 + t^2)^{2n + 2}} \sin \left [4(n + 1) \tan^{-1} \left (\frac{1}{t} \right ) \right ],$$ and $$\mathcal{L}^{-1} \{e^{-x} \} (t)= \delta (t - 1),$$ where $\delta (x)$ is the Dirac delta function, then \begin{align} I_n &= 4\int_0^\infty x^{4n + 3} \sin x \cdot e^{-x} \, dx\\ &= 4\int_0^\infty \mathcal{L} \{x^{4n + 3} \sin x\} (t) \cdot \mathcal{L}^{-1} \{e^{-x} \} (t) \, dt\\ &= 4(4n + 3)! \int_0^\infty \frac{1}{(1 + t^2)^{2n + 2}} \sin \left [4(n + 1) \tan^{-1} \left (\frac{1}{t} \right ) \right ] \cdot \delta (t - 1) \, dt\\ &= \frac{4(4n + 3)!}{2^{2n + 2}} \sin [4(n + 1) \tan^{-1} (1)]\\ &= \frac{(4n + 3)!}{4^n} \sin ((n + 1)\pi)\\ &= 0, \end{align} as required to show.

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A direct approach

Here is an approach where the results of the above Laplace transform and its inverse are not quoted in advance.

From $$I_n = 4 \int_0^\infty e^{-x} x^{4n + 3} \sin x \, dx,$$ we re-write this as $$I_n = -4 \operatorname{Im} \int_0^\infty x^{4n + 3} e^{-(1 + i)x} \, dx.$$ Integrating by parts $(4n + 3)$ times gives \begin{align} I_n &=-4 \, \operatorname{Im} \left [\frac{(-1)^{4n + 3} (4n + 3)!}{(1 + i)^{4n + 3}} \right ] \int_0^\infty e^{-(1 + i)x} \, dx\\ &= 4 \, \operatorname{Im} \left [\frac{(-1)^{4n + 4} (4n + 3)!}{(1 + i)^{4n + 4}} \right ]\\ &= 4(4n + 3)! \operatorname{Im} \left [\frac{1}{(1 + i)^{4n + 4}} \right ]\\ &= 0, \end{align} where the last line is due to the fact that $$\frac{1}{(1 + i)^{4n + 4}} = \frac{1}{2^{2n + 2}} \left [\cos ((n + 1)\pi) + i \sin ((n + 1)\pi) \right ].$$