Integral of $R(R^2+y^2)^{-3/2}$ with respect to $y$

Question

$$\int_0^\infty \frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}dy$$

The indefinite integral seems to be $$\frac{-R}{\sqrt{R^2+y^2}}+C$$

$R$ is a constant

Answer 1

We assume that $R$ is positive.

Let $y=R\tan\theta$ or equivalently $\theta=\arctan(y/R)$. As $y$ ranges from $0$ to $\infty$, the number $\theta$ ranges from $0$ to $\pi/2$.

Note that $\sqrt{R^2+y^2}\left(R^2+y^2\right)= R^3 \sec^3\theta$ and $dy=R\sec^2\theta\,d\theta$. We leave the rest to you.

Answer 2

Using $y=Rx$, then $x=\tan(\theta)$ we get $$ \begin{align} \int_0^\infty\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y &=\frac1R\int_0^\infty\frac{\mathrm{d}x}{\sqrt{1+x^2}\left(1+x^2\right)}\\ &=\frac1R\int_0^{\pi/2}\cos(\theta)\,\mathrm{d}\theta\\ &=\frac1R \end{align} $$ It seems that you have miscomputed the primitive.

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Computing the Primitive

Combining the previous substitutions with $y=R\tan(\theta)$, we get $$ \begin{align} \int\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y &=\frac1R\int\cos(\theta)\,\mathrm{d}\theta\\ &=\frac1R\sin(\theta)+C\\ &=\frac1R\frac{y}{\sqrt{R^2+y^2}}+C \end{align} $$

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\overbrace{\color{#66f}{\large\int_{0}^{\infty}{R \over \pars{R^{2} + y^{2}}^{3/2}}\,\dd y}} ^{\ds{y = \verts{R}\sinh\pars{\theta}}}\ =\ {R \over \verts{R}^{2}}\int_{0}^{\infty}\sech^{2}\pars{\theta}\,\dd\theta =\left.{1 \over R}\,\tanh\pars{\theta} \,\right\vert_{\,\theta\ =\ 0}^{\,\theta\ \to\ \infty} \\[3mm]&=\color{#66f}{\large{1 \over R}} \end{align}