$\DeclareMathOperator{\arcsinh}{arcsinh} \DeclareMathOperator{\sgn}{sgn} \newcommand{\dd}{\mathop{}\!\mathrm{d}}$

My integral follows: $$\int\limits_0^n\left(\int\limits_0^n \dfrac{1}{\sqrt{x^2+y^2}+1}\dd x\right)\dd y.$$ I attempted the following: $$(\text{integral above})=\int\limits_0^n\left(\int_0^1\dfrac{1}{|y|\sqrt{t^2+1}+1}y\dd t\right)\dd y,$$ by substituting $t=\frac{x}{y}$, whence $\dd x=y\dd t$. Let's have a look at that internal integral. The root suggests a hyperbolic substitution $t=\sinh z$, whence $\dd t=\cosh z\dd z$. That leads to: $$\int\limits_0^{\arcsinh(1)}\dfrac{1}{|y|\cosh z+1}y\cosh z\dd z=\sgn(y)\left(\int\limits_0^1\dfrac{|y|\cosh(z)+1-1}{|y|\cosh(z)+1}\dd z\right)=$$ $$=\sgn(y)\arcsinh\Big(\frac{x}{y}\Big)\Big|_0^n-\sgn(y)\int\limits_0^{\arcsinh(1)}\dfrac{1}{|y|\cosh(z)+1}\dd t.$$ That is where I got stuck. The first term clearly evaluates to $\sgn(y)\arcsinh(\frac{n}{y})$, but then how do I integrate it in $\dd y$? And how do I integrate the second term? Is there an easier way to do this?

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    $\begingroup$ Try x = rcos(t), y = rsin(t) with dxdy = rdrdt and some suitable limits. $\endgroup$ – Paul Sep 18 '14 at 10:45
  • $\begingroup$ I agree. The limits seem to be and (0, n/cos(t)) for r and (0, $pi$/2) for t. $\endgroup$ – A. Bellmunt Sep 18 '14 at 11:05
  • $\begingroup$ Since the actual goal of this is to prove that the measure with the integrated function as density relative to the 2D Lebesgue measure is sigma-finite, using that on balls of radius n solves the problem. Anyway to find this integral, what limits would I need? Radius from 0 to $\frac{n}{\cos(\theta)}$ and angle from 0 to $\frac{\pi}{2}$, right? $\endgroup$ – MickG Sep 18 '14 at 11:12
  • $\begingroup$ OK we commented in the same time :). $\endgroup$ – MickG Sep 18 '14 at 11:13
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    $\begingroup$ When you turn an integral over a square $[0,n]\times[0,n]$ into polar coordinates, you need to break the polar coordinate integral into two pieces, with the break at $\theta=\pi/4$. (Actually, by the symmetry here, you only need to do the piece from $0$ to $\pi/4$.) $\endgroup$ – Barry Cipra Sep 18 '14 at 11:51

This was once an edit to the question, but it should be an answer, with opportune completions.


Taking the comments into account, here is another approach.

Using polar coordinates, the integral becomes: $$\int\limits_0^{\frac{\pi}{2}}\left(\int\limits_?^?\dfrac{r}{1+r}\mathrm{d}r\right)\mathrm{d}\theta.$$ By symmetry, we turn this into twice the integral with theta stopping at $\frac{\pi}{4}$. With that, the inner limits are 0 and $\frac{n}{\cos(\theta)}$. In the other half, the cosine in the limit would be a sine. Thus the integral is: $$2\cdot\int\limits_0^{\frac{\pi}{4}}\left(\int\limits_0^{\frac{n}{\cos(\theta)}}\dfrac{r}{1+r}\mathrm{d}r\right)\mathrm{d}\theta=2\cdot\int\limits_0^{\frac{\pi}{4}}\left(r-\log(1+r)\right)\Big|_0^{\frac{n}{\cos(\theta)}}\mathrm{d}\theta.$$ The evaluation of the integrand yields: $$\frac{n}{\cos(\theta)}-\log(1+\frac{n}{\cos(\theta)}).$$

For completeness's sake, the limits for the other part are 0 and $\frac{n}{\sin(\theta)}$, making it the same as this one, as is obvious in cartesian coordinates by a symmetry of the function.

Now we need but to complete the evaluation of that integral. For that, I need the primitive of $\frac{1}{\cos\theta}$:

\begin{align*} \int\frac{1}{\cos\theta}d\theta={}&\frac{\theta}{\cos\theta}-\int\frac{\theta}{-\cos^2\theta}(-\sin\theta) d\theta=\frac{\theta}{\cos\theta}-\int\theta\frac{d}{d\theta}(\tan\theta)\sin\theta d\theta={} \\ {}={}&\frac{\theta}{\cos\theta}-\theta\tan\theta\sin\theta+\int\theta\tan\theta(\theta\cos\theta+\sin\theta)d\theta={} \\ {}={}&\frac{\theta}{\cos\theta}-\theta\tan\theta\sin\theta+\int\left(\theta\sin\theta+\tan\theta\sin\theta\right)d\theta. \end{align*}

The first term in that last integral integrates to $-\theta\cos\theta+\sin\theta$. The second term integrates to $-\tan\theta\cos\theta+\int\frac{\cos\theta}{\cos^2\theta}d\theta$, oh dear, all this work for nothing… If I substitute $t=\cos\theta$, the integral becomes $\frac{1}{t\sqrt{1-t^2}}dt$. We substitute $u=\sqrt{1-t^2}$ and get $t=\sqrt{1-u^2}$ and $du=-\frac{t}{\sqrt{1-t^2}}dt$ or $dt=-\frac{u}{\sqrt{1-u^2}}du$, so the new integral is $-\frac{1}{u\sqrt{1-u^2}}\frac{u}{\sqrt{1-u^2}}du=-\frac{1}{1-u^2}du=-\frac12(\frac{1}{1-u}+\frac{1}{1+u})du$, so this evaluates to $-\frac12\log(1-u)-\frac12\log(1+u)=-\frac12\log(1-u^2)$, and going back to $t$ this is $-\frac12\log t^2$, and back into $\theta$ we get $-\frac12\log(\cos^2\theta)=-\log(\cos\theta)$. Which is evidently wrong, so what am I doing wrong? I won't even attempt to set the other term straight.


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