Classic integral

Calculus (Integrals, Series)
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Tolaso J Kos
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Classic integral

#1

Post by Tolaso J Kos »

Evaluate the integral:

$$\int_{-\infty}^{\infty} \frac{dx}{\left(1+x^2 \right)^k}, \quad k \in \mathbb{N}$$
Imagination is much more important than knowledge.
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Grigorios Kostakos
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Re: Classic integral

#2

Post by Grigorios Kostakos »

We use the identities: \begin{align*}
{\rm{B}}(a,b)&=\int_{0}^{+\infty}{\frac{x^{a-1}}{(1+x)^{a+b}}dx}\,,\;a>0,b>0&\quad(1)\\
{\rm{B}}(a,b)&=\frac{\Gamma(a)\,\Gamma(b)}{\Gamma(a+b)}\,,\;a>0,b>0&\quad(2)\\
\Gamma\bigl({n+\tfrac{1}{2}}\bigr)&=\frac{(2n)!\sqrt{\pi}}{4^{n}n!}\,,\;n\in{\mathbb{N}}&\quad(3)
\end{align*} So \begin{align*}
\displaystyle \int_{-\infty}^{+\infty} {\frac{dx}{(1+x^2)^k}}&=\int_{-\infty}^{0}{ \frac{dx}{(1+x^2)^k}}+\int_{0}^{+\infty} {\frac{dx}{(1+x^2)^k}}\\
&\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,-x}\\
{-dt\,=\,dx}\\
\end{subarray}}\,\int_{0}^{+\infty}{\frac{dt}{(1+t^2)^k}}+\int_{0}^{+\infty}{\frac{dx}{(1+x^2)^k}}\\
&=2\int_{0}^{+\infty}{\frac{dx}{(1+x^2)^k}}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,x^2}\\
{\frac{1}{2}t^{-\frac{1}{2}}dt\,=\,dx}\\
\end{subarray}}\,\int_{0}^{+\infty}{\frac{t^{-\frac{1}{2}}}{(1+t)^k}dt}\\
&=\int_{0}^{+\infty}{\frac{t^{\frac{1}{2}-1}}{(1+t)^{k-\frac{1}{2}+\frac{1}{2}}}dt}\\
&\stackrel{(1)}{=}{\rm{B}}\bigl({\tfrac{1}{2},k-\tfrac{1}{2}}\bigr)\\
&\stackrel{(2)}{=}\frac{\Gamma\bigl({\tfrac{1}{2}}\bigr)\Gamma\bigl({k-\tfrac{1}{2}}\bigr)}{\Gamma(k)}\\
&=\frac{\sqrt{\pi}}{\Gamma(k)}\Gamma\bigl({(k-1)+\tfrac{1}{2}}\bigr)\\
&\stackrel{(3)}{=}\frac{\sqrt{\pi}}{(k-1)!}\frac{(2k-2)!\sqrt{\pi}}{4^{k-1}(k-1)!}\\
&=\frac{\sqrt{\pi}}{(k-1)!}\frac{2(2k-3)!\sqrt{\pi}}{4^{k-1}(k-2)!}\\
&=\frac{8\pi(2k-3)!}{4^k (k-1)!(k-2)!}\,,\quad k\geqslant2\,.
\end{align*}
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Re: Classic integral

#3

Post by Papapetros Vaggelis »

2nd solution :

If \(\displaystyle{k\in\mathbb{N}}\) , then \(\displaystyle{x\in\mathbb{R}\implies \dfrac{1}{\left(1+x^2\right)^{k}}\leq \dfrac{1}{1+x^2}}\)

and \(\displaystyle{\int_{-\infty}^{\infty}\dfrac{1}{1+x^2}\,\rm{dx}=\left[\arctan\,x\right]_{-\infty}^{\infty}=\pi<\infty}\) ,

so \(\displaystyle{\int_{-\infty}^{\infty}\dfrac{1}{\left(1+x^2\right)^{k}}\,\rm{dx}<\infty\,\forall\,k\in\mathbb{N}}\) .

Also, \(\displaystyle{\int_{-\infty}^{\infty}\dfrac{1}{\left(1+x^2\right)^{k}}\,\rm{dx}=2\,\int_{0}^{\infty}\dfrac{1}{\left(1+x^2\right)^{k}}\,\rm{dx}}\) .

We define the real sequence \(\displaystyle{\left(a_{k}\right)_{k\in\mathbb{N}}}\) by \(\displaystyle{a_{k}=\int_{0}^{\infty}\dfrac{1}{\left(1+x^2\right)^{k}}\,\rm{dx}}\) .

We know that \(\displaystyle{a_1=\dfrac{\pi}{2}}\) and for each \(\displaystyle{k\in\mathbb{N}\,,k\geq 2}\) holds :

\(\displaystyle{\begin{aligned}a_{k}&=\int_{0}^{\infty}\dfrac{1}{\left(1+x^2\right)^{k}}\,\rm{dx}\\&=\left[\dfrac{x}{\left(1+x^2\right)^{k}}\right]_{0}^{\infty}-\int_{0}^{\infty}x\,\dfrac{-2\,k\,x\,\left(1+x^2\right)^{k-1}}{\left(1+x^2\right)^{2\,k}}\,\rm{dx}\\&=2\,k\,\int_{0}^{\infty}\dfrac{\left(1+x^2\right)-1}{\left(1+x^2\right)^{k+1}}\,\rm{dx}\\&=2\,k\,a_{k}-2\,k\,a_{k+1}\\&\implies a_{k+1}=\dfrac{2\,k-1}{2\,k}\,a_{k}\,\,(I)\end{aligned}}\)

Therefore, we guess that :

\(\displaystyle{\begin{aligned} a_{k+1}&=\dfrac{2\,k-1}{2\,k}\,a_{k}\\&=\dfrac{2\,k-1}{2\,k}\,\dfrac{2\,k-3}{2\,(k-1)}\,a_{k-1}\\&=\dfrac{2\,k-1}{2\,k}\,\dfrac{2\,k-3}{2\,(k-1)}\,\dfrac{2\,k-5}{2\,(k-2)}\,a_{k-2}\\&=...=\dfrac{1}{2^{k}\,\left(k-1\right)!}\,\prod_{n=1}^{k}\left[2\,k-\left(2\,n-1\right)\right]\,a_{1}\\&=\dfrac{\pi}{2^{k+1}\,k!}\,\prod_{n=1}^{k}\left[2\,k-\left(2\,n-1\right)\right]\,,k\in\mathbb{N}\end{aligned}}\)

We' ll prove the above equality inductively.

\(\displaystyle{\bullet\,\,k=2 :}\)

\(\displaystyle{a_{2}\stackrel{(I)}{=}\dfrac{a_{1}}{2}=\dfrac{\pi}{4}}\) and

\(\displaystyle{\dfrac{\pi}{2^{2}\,1!}\,\prod_{n=1}^{1}\left[2\,k-\left(2\,n-1\right)\right]=\dfrac{\pi}{4}\,(2-1)=\dfrac{\pi}{4}=a_{2}}\)

We assume that \(\displaystyle{a_{k+1}=\dfrac{\pi}{2^{k+1}\,k!}\,\prod_{n=1}^{k}\left[2\,k-\left(2\,n-1\right)\right]\,\,(II)\,,k\geq 2}\)

\(\displaystyle{\bullet\,\,k\to k+1}\)

\(\displaystyle{\begin{aligned} a_{k+2}&=\dfrac{2\,k+1}{2\,(k+1)}\,a_{k+1}\\&=\dfrac{2\,k+1}{2\,(k+1)}\,\dfrac{\pi}{2^{k+1}\,k!}\,\prod_{n=1}^{k}\left[2\,k-\left(2\,n-1\right)\right]\\&=\dfrac{\pi}{2^{k+2}\,(k+1)!}\,\prod_{n=1}^{k+1}\left[2\,(k+1)-\left(2\,n-1\right)\right]\end{aligned}}\)

So, \(\displaystyle{a_{k}=\dfrac{8\,\pi}{2^{k+3}\,\left(k-1\right)!}\,\prod_{n=1}^{k-1}\left[2\,k-\left(2\,n+1\right)\right]\,,k\geq 2\,,a_1=\dfrac{\pi}{2}}\)

and then \(\displaystyle{\int_{-\infty}^{\infty}\dfrac{1}{1+x^2}\,\rm{dx}=\pi}\) ,

\(\displaystyle{\int_{-\infty}^{\infty}\dfrac{1}{\left(1+x^2\right)^{k}}\,\rm{dx}=\dfrac{8\,\pi}{2^{k+2}\,\left(k-1\right)!}\,\prod_{n=1}^{k-1}\left[2\,k-\left(2\,n+1\right)\right]\,,k\geq 2}\)
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