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Prove that $$\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right) \, dy,\: n\in \mathbb{N}$$

My try

  • $n=0$ then $\frac{\sin x}{x}=\frac{1}{x} \int_0^x \cos y\,dy$ (true)
  • Assuming $n=k$ (true)
  • Prove $n=k+1$
Michael Hardy
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Iloveyou
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  • What do you want to ask? – Raven Feb 26 '14 at 04:02
  • Suggestion that I'm not certain of (I'll come back in a couple hours and try to make it a solution if there isn't one by then): Subtract the right-hand side to the left and try to reverse the usual steps in an integration by parts. – Addem Feb 26 '14 at 04:04

3 Answers3

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Here is how you advance. We need the following identity

$$ \frac{d^n}{dx^n}\cos(tx) = t^n\cos(tx+n\pi/2) $$

which is easy to prove. We start with the integral representation of $\frac{\sin x}{x}$

$$ \frac{\sin x}{x} = \int_{0}^{1} \cos (tx) dt \implies \left(\frac{\sin x}{x}\right)^{(n)} = \int_{0}^{1} t^n\cos(tx+n\pi/2) dt, $$

which follows from the above identity. Now, make the change of variables $xt=y$ which yields the desired result

$$ \left(\frac{\sin x}{x}\right)^{(n)} = \frac{1}{x^{n+1}}\int_{0}^{x} y^n\cos(y+n\pi/2) dy. $$

Mhenni Benghorbal
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$\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{d}{dx}\frac{d^{n-1}}{dx^{n-1}}\left(\frac{\sin x}{x}\right)$

$=\frac{d}{dx}\frac{1}{x^{n}}\int_{0}^{x} y^{n-1}\cos\left(y+\frac{(n-1)\pi}{2}\right)dy$ (inductive assumption)

$=\frac{-n}{x^{n+1}}\int_{0}^{x} y^{n-1}\cos\left(y+\frac{(n-1)\pi}{2}\right)dy+\frac{1}{x^n}\frac{d}{dx}\int_{0}^{x} y^{n-1}\cos\left(y+\frac{(n-1)\pi}{2}\right)dy$ (Product Rule)

$=\frac{-n}{x^{n+1}}\int_{0}^{x} y^{n-1}\cos\left(y+\frac{(n-1)\pi}{2}\right)dy+\frac{1}{x^n}x^{n-1}\cos\left(x+\frac{(n-1)\pi}{2}\right)$ (Fundamental Theorem of Calculus)

$=\frac{-n}{x^{n+1}}\left(|\frac{1}{n}y^n\cos\left(y+\frac{(n-1)\pi}{2}\right)|_0^x+\int_{0}^{x} \frac{1}{n}y^{n}\sin\left(y+\frac{(n-1)\pi}{2}\right)dy\right)+\frac{1}{x^n}x^{n-1}\cos\left(x+\frac{(n-1)\pi}{2}\right)$ (integration by parts)

$=\frac{-1}{x^{n+1}}\left(x^n\sin\left(x+\frac{n\pi}{2}\right)-\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right)dy\right)+\frac{1}{x^n}x^{n-1}\sin\left(x+\frac{n\pi}{2}\right)$ (trigonometric identities)

$=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right)dy$ (first and last terms cancel)

Andrew
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Let $f_n(x)=\int_0^x y^n \cos(y+n\pi/2) dy$. Then, using integration by parts,

$$\int_0^z \frac{1}{x^{n+1}} f_n(x) dx = -\frac{1}{nz^n} f_n(z) + \frac{1}{n} \sin(z+n\pi/2).$$

Now, again using integration by parts, this time on $f_n(z)$, it follows that the right hand side equals

$$\frac{1}{z^n} \int_0^z y^{n-1} \sin(y+n\pi/2)dy.$$

Note that $\sin(y+n\pi/2) = \cos(y+(n-1)\pi/2)$, which follows from $\sin(a+b)=\sin a \cos b + \sin b \cos a$.

Hence,

$$\int_0^z \frac{1}{x^{n+1}} f_n(x) dx = \frac{1}{z^n} f_{n-1}(z).$$

Now use your induction trick.

JPi
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