4
$\begingroup$

Where f(x) is an even function,and also is a periodic functions,the period is $\pi$,and $n,m\in\Bbb N$,and n+m is odd number. When n+m is even, I already have a solution.But when n+m is odd, I don't know how to solve it.I'd appreciate it if someone could help me with it.

When n+m is even,my answer is as follows:

If f (x) is an even function, and the period is $\pi$,we have: $$\int_{0}^\infty f(x)\frac{\sin^nx}{x^m}dx=\int_{0}^\frac{\pi}{2}f(x)g_m(x)\sin^nxdx \qquad (1)$$

$n,m\in\Bbb N$, Where the n+m is an even,and $g_m(x)$ in (1) is as follows: $$g_m(x)=\begin{cases}\frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{dx^{m-1}}\left(\csc x\right),& \text{for n is odd and}\\[2ex] \frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{dx^{m-1}}\left(\cot x\right),& \text{ for n is even .} \end{cases}$$ —————————————————————————————————————————————————— Proof: \begin{align} \int_{0}^\infty f(x)\frac{\sin^nx}{x^m}dx&=\sum_{k=0}^\infty\int_{k\pi}^{(2k+1)\frac{\pi}{2}}f(x)\left(\frac{\sin ^nx}{x^m}\right)dx+\sum_{k=1}^\infty\int_{(2k-1)\frac{\pi}{2}}^{k\pi}f(x)\left(\frac{\sin^n x}{x^m}\right)dx\\ &=\sum_{k=0}^\infty\int_{0}^{\frac{\pi}{2}}f(x+k\pi)\left(\frac{\sin^n (x+k\pi)}{(x+k\pi)^m}\right)dx+\sum_{k=1}^\infty\int_{-\frac{\pi}{2}}^{0}f(x+k\pi)\left(\frac{\sin^n (x+k\pi)}{(x+k\pi)^m}\right)dx\\ &=\sum_{k=0}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(x)\left(\frac{\sin^n x}{(x+k\pi)^m}\right)dx+\sum_{k=1}^\infty(-1)^{nk+n+m}\int_{0}^{\frac{\pi}{2}}f(-x)\left(\frac{\sin^n x}{(x-k\pi)^m}\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nx\left(\frac{1}{x^m}+\sum_{k=1}^\infty(-1)^{nk}\left[\frac{1}{(x+k\pi)^m}+\frac{(-1)^{n+m}}{(x-k\pi)^m}\right]\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nxg_m(x)dx \end{align} When n+m is an even,and we know by the Fourier series \begin{align} \csc x&=\frac{1}{x}+\sum_{k=1}^\infty(-1)^k\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)\\ \end{align} and \begin{align} \cot x&=\frac{1}{x}+\sum_{k=1}^\infty\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right) \end{align} Take the m-1 order derivative,thus we obtain $g_m(x)$. —————————————————————————————————————————————————— Example: \begin{align} (1.)\qquad\int_{0}^{\infty}\frac{\sin^3x}{x}dx&=\int_{0}^{\frac{\pi}{2}}\sin^2xg_1(x)\sin xdx\\ &=\int_{0}^{\frac{\pi}{2}}\sin^2x\frac{1}{\sin x}\sin xdx\\ &=\int_{0}^{\frac{\pi}{2}}\sin^2xdx\\ &=\frac{\pi}{4}\\ \end{align} \begin{align} (2.) \int_{0}^{\infty}(1+\cos^2x)\frac{\sin^2x}{x^2}dx &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)g_2(x)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(-\frac{d}{dx}\cot x\right)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(\frac{1}{\sin^2x}\right)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)dx\\ &=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}\\ \end{align} \begin{align} (3.) \int_{0}^{\infty}\frac{1}{(1+\cos^2x)}\frac{\sin^3x}{x^3}dx &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}g_3(x)dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\left(\frac{1}{2}\frac{d^2}{dx^2}(\csc x)\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\frac{(1+\cos^2x)}{2\sin^3x}dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}dx=\frac{\pi}{4}\\ \end{align}

$\endgroup$
1
  • $\begingroup$ When n=0,That's a perfect formula, but suppose the integral converges. $\endgroup$
    – JamesJ
    Aug 28, 2017 at 8:57

1 Answer 1

5
$\begingroup$

There is an excellent result related with this integral, enjoy!

$$I=\int_{0}^{\infty }x^{p}\ \left ( \frac{\sin(x)}{x} \right )^ndx\ \ \ \ \ \ , n=1,2,3,\dots , \ \ \ 0\geq p\geq -1,\\ \\ \\ I=\frac{\pi }{2(2i)^{n}\Gamma (n-p)}\sum_{m=0}^{n }(-1)^{n-m}\frac{n!}{m!(n-m)!}\left | n-2m \right |^{n-p-1}\left ( \frac{1}{\sin(\frac{n-p+1}{2})\pi }-\frac{\operatorname{sgn}(n-2m)}{\sin(\frac{n-p}{2})\pi }i \right ).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.