I need to find the definite integral of $\int(1+x^2)^{-4}~dx$ from $0$ to $\infty$ .
I rewrite this as $\dfrac{1}{(1+x^2)^4}$ .
The $\dfrac{1}{1+x^2}$ part, from $0$ to $\infty$ , seems easy enough: $\arctan(\infty)-\arctan(0)$ which gives $\dfrac{\pi}{2}$ .
How do I deal with the "$^4$" part of the equation.
I know the answer is $\dfrac{5\pi}{32}$ . What steps am I missing?
Thank you.
On
Consider the sub. $x=\tan\theta$ you will get a function of $\cos \theta$ with even power, try to reduce power by double angle rule
On
You can't deal with the integrand in that manner -- the 4th power and the $1/(1+x^2)$ must be dealt with 'together'. Denote your integral by $I$ and use a substitution of $x=\tan \theta$ and the identity $\sec ^2 \theta = 1+ \tan ^2 \theta$ to reduce the integrand to $\cos ^6\theta$.
From there, you have multiple ways in. One of which is to let $J_n = \int_0^{\pi/2} \cos^{2n}\theta d\theta$ and derive an reduction formula via parts by writing the integrand as $\cos^{2n-1}\theta \cdot \sin \theta$.
Alternatively, use Alan's approach.
EDIT: If you want a more fancy way that abuses symmetry, http://www.thestudentroom.co.uk/showthread.php?t=1608009&p=30894842#post30894842
On
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{t \equiv {1 \over 1 + x^{2}}\quad\iff\quad x = \pars{{1 \over t} - 1}^{1/2}}$ \begin{align} \int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}^{4}}&= \int_{1}^{0}t^{4}\,\half\,\pars{1 - t \over t}^{-1/2}\pars{-\,{1 \over t^{2}}}\,\dd t =\half\int_{0}^{1}t^{5/2}\pars{1 - t}^{-1/2}\,\dd t =\half\,{\rm B}\pars{{7 \over 2},\half} \end{align} where ${\rm B}\pars{x,y}$ is the Beta Function which satisfies: $$ {\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}\,, \qquad\Gamma\pars{z} :\ Gamma\ Function $$ \begin{align} \color{#00f}{\large\int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}^{4}}}&= \half\,{\overbrace{\Gamma\pars{7/2}}^{\ds{15\root{\pi} \over 8}}\ \overbrace{\Gamma\pars{1/2}}^{\ds{\root{\pi}}} \over \underbrace{\Gamma\pars{7/2 + 1/2}}_{3! = 6}} =\color{#00f}{\large{5\pi \over 32}} \approx 0.4909 \end{align}
On
Let $\forall n\in \mathbb{N}, I_n(x):=\int_0^x \frac{dt}{(t^2+1)^n}$ by integration by parts we get : $$ I_n(x)=\frac{x}{(1+x^2)^n}+2n(I_n(x)-I_{n+1}(x)) $$ Thus, $$ I_{n+1}(x)=\frac{2n-1}{2n}I_n(x)+\frac{1}{2n}\frac{x}{(1+x^2)^n} $$ Note that $I_1=arctan (x)$.
Therefore, $$ I_{4}(x)=\frac{1}6\frac{x}{(1+x^2)^3}+\frac{5}{24}\frac{x}{(1+x^2)^2}+\frac{5}{16}\frac{x}{(1+x^2)}+\frac{5}{16}arctan x $$ Take the limit as $x\rightarrow +\infty$ we get the result $\frac{5\pi}{32}$.
Substitute $x=\tan{t}$. Then, $$\int_0^\infty \frac{dx}{(x^2+1)^4} dx=\int_0^\frac{\pi}{2} \cos^6{t} dt=\int_0^{\frac{\pi}{2}} (\frac{1+\cos{2t}}{2})^3 dt.$$ By repeated use of formula $\cos^2{t}=\frac{1+\cos{2t}}{2}$, it is easy to get result $\frac{5\pi}{32}$.