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$$\int_{-b}^b\int_{-a}^a\frac{1}{(x^2+y^2+h^2)^{3/2}}dxdy$$

First I put x=$\sqrt{y^2+h^2}tan\theta$ and arrive at: $$\int_{-b}^b\frac{2a}{(y^2+h^2)(y^2+h^2+a^2)^{1/2}}dy$$

Can you please tell me what to do after this?

Beck Humh
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  • What did you try? Did you try to let $u=y/\sqrt{y^2+h^2+a^2}$? Also, as you have written your first integral the limits for $x$ is $-b$ and $b$. If that is not intended, then you might want to update your question with the different order. – mickep Apr 08 '17 at 18:19

2 Answers2

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\pars{~\mbox{with}\ a\,,\ b > 0~}}$ and $\ds{\pars{p \equiv {a \over \verts{h}} > 0\,,\ q \equiv {b \over \verts{h}} > 0}}$:

$$ \int_{-a}^{a}\int_{-b}^{b}{\dd x\,\dd y \over \pars{x^{2} + y^{2} + h^{2}}^{3/2}} = {4 \over \verts{h}}\bbox[5px,#ffd]{\ds{\int_{0}^{q}\int_{0}^{p}{\dd x\,\dd y \over \pars{x^{2} + y^{2} + 1}^{3/2}}}} $$

\begin{align} &\bbox[5px,#ffd]{\ds{\int_{0}^{q}\int_{0}^{p}{\dd x\,\dd y \over \pars{x^{2} + y^{2} + 1}^{3/2}}}} \,\,\,\stackrel{y\ \mapsto\ {1/y}}{=}\,\,\, \int_{0}^{p}\int_{1/q}^{\infty}{y\,\dd y \over \bracks{\pars{x^{2} + 1}y^{2} + 1}^{3/2}}\,\dd x \\[5mm] \stackrel{y^{2}\ \mapsto\ y}{=} &\ {1 \over 2}\int_{0}^{p}\int_{1/q^{2}}^{\infty}{\dd y \over \bracks{\pars{x^{2} + 1}y + 1}^{3/2}}\,\dd x = -\int_{0}^{p}{1 \over x^{2} + 1}\left.% {1 \over \bracks{\pars{x^{2} + 1}y + 1}^{1/2}} \right\vert_{\ y\ =\ 1/q^{2}}^{\ y\ \to\ \infty}\,\dd x \\[5mm] = &\ q\int_{0}^{p}{1 \over x^{2} + 1} {1 \over \bracks{\pars{x^{2} + 1} + q^{2}}^{1/2}}\,\dd x \,\,\,\stackrel{x\ \mapsto\ {1/x}}{=}\,\,\, q\int_{1/p}^{\infty} {x\,\dd x \over \pars{x^{2} + 1}\bracks{\pars{q^{2} + 1}x^{2} + 1}^{1/2}} \\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,& {1 \over 2}\,q\int_{1/p^{2}}^{\infty} {\dd x \over \pars{x + 1}\bracks{\pars{q^{2} + 1}x + 1}^{1/2}} \end{align}

With the change of variable $\ds{x \equiv {t^{2} - 1 \over q^{2} + 1}}$:

\begin{align} &\bbox[5px,#ffd]{\ds{\int_{0}^{q}\int_{0}^{p}{\dd x\,\dd y \over \pars{x^{2} + y^{2} + 1}^{3/2}}}} = q\int_{\root{p^{2} + q^{2} + 1}/p}^{\infty}{\dd t \over t^{2} + q^{2}} = {\pi \over 2} - \arctan\pars{\root{p^{2} + q^{2} + 1} \over pq} \\[5mm] = &\ \arctan\pars{pq \over \root{p^{2} + q^{2} + 1}} \end{align}

$$ \bbx{\ds{\int_{-a}^{a}\int_{-b}^{b}{\dd x\,\dd y \over \pars{x^{2} + y^{2} + h^{2}}^{3/2}} = {4 \over h}\,\arctan\pars{ab/h \over \root{a^{2} + b^{2} + h^{2}}}}} $$
Felix Marin
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  • Now that after 7 years I am doing something worthwhile and not wasting my time, I had been stuck in this problem for 2 days and I felt very discouraged. I know this is too much but thank you. You saved me from giving up. – Beck Humh Apr 09 '17 at 15:52
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    @BeckHumh Thanks. I'm glad it was useful. Something I always check at the beginning is the change $x \mapsto 1/x$. It yields many surprises. – Felix Marin Apr 09 '17 at 22:09
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If we substitute $$y=\sqrt{a^2+h^2}\tan u$$ $$dy=\sqrt{a^2+h^2}\sec^2u \text{ }du$$ We get the integral $$\int \frac{2a(\sqrt{a^2+h^2}\sec^2u)}{((a^2+h^2)\tan^2u+h^2)((a^2+h^2)\tan^2u+a^2+h^2)^{1/2}}du$$ Remembering that $\tan^2x+1=\sec^2x$, we get $$\int \frac{2a\sqrt{a^2+h^2}\sec^2u}{((a^2+h^2)\sec^2u -h^2)((a^2+h^2)\sec^2u)^{1/2}}du$$ Now you can simplify quite a bit.

Hope this helps!

Isaac Browne
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