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

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

Edit:

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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