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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?

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  • $\begingroup$ Mathematical induction and integration by parts. $\endgroup$
    – azif00
    Jul 10, 2019 at 21:48
  • $\begingroup$ You missed a factor $4$ in the second formula (from $dx=4u^3du$). It doesn't matter much, since it still going to be $0$. $\endgroup$ Jul 10, 2019 at 23:17
  • $\begingroup$ @AdamLatosiński but that won't affect his final outcome, if his claim is indeed correct. $\endgroup$
    – The Count
    Jul 10, 2019 at 23:19
  • $\begingroup$ thanks I edited $\endgroup$
    – sadeem
    Jul 11, 2019 at 19:28

5 Answers 5

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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 $$ F_n(1-i) = \frac{(4n+3)!}{(1-i)^{4n+4}} = \frac{(4n+3)!}{(-4)^{n+1}} \in \mathbb R, $$ $I(n) = 0$.

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  • $\begingroup$ But proving that Laplace transform requires the gamma function, which is just partial fraction in disguise.. It looks easier only because here the results were given, or am I wrong? $\endgroup$
    – Zacky
    Jul 11, 2019 at 7:11
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    $\begingroup$ @Zacky That's true, but 1) integrating the gamma integral by parts is way easier than the one in the OP, and 2) finding ways to outsource your mathematics to an integral table is an important skill in calculus. $\endgroup$ Jul 11, 2019 at 15:08
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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 $4{{A}_{4n+3}}=0$.

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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 $$I(n) = {\rm Im} \Big(\int_0^\infty 4u^{4n+3} e^{iu} e^{-u} du \Big) = {\rm Im} \Big(4\int_0^\infty u^{4n+3} e^{-u(1-i)} du \Big) = 0$$

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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).

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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.


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 $$\frac{1}{(1 + i)^{4n + 4}} = \frac{1}{2^{2n + 2}} \left [\cos ((n + 1)\pi) + i \sin ((n + 1)\pi) \right ].$$

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    $\begingroup$ $\newcommand{\L}{\mathcal{L}}$Isn't $\int_0^\infty \left(x^{4n+3}\sin x\right) e^{-x}\, dx$ just $\L\left(x^{4n+3}\sin x \right)(s)\big{\vert}_{s=1}$? So if you can quote the Laplace transform of $x^{4n+3}\sin x$, you can essentially just jump straight to quoting the value of $\int_0^\infty \left(x^{4n+3}\sin x\right) e^{-x}\, dx$ (and also don't need to use special properties). $\endgroup$ Jul 12, 2019 at 10:26
  • $\begingroup$ @ Minus One-Twelfth - Indeed, so I have updated my answer by providing a direct approach that does not rely on quoting the Laplace transform of $x^{4n + 3} \sin x$. $\endgroup$
    – omegadot
    Jul 13, 2019 at 1:47

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