My coursebook on integral calculus has a question:

  1. By changing into polar coordinates, show that: $$ \int_0^a\int_y^a \frac{1}{x^2+y^2} dx dy = \frac{\pi a}{4} $$

This doesn't look good to me. When we take $y$ approaching $0$ (our lower bound on $y$), since value of $y$ is a lower limit for $x$, $x$ too approaches $0$ on the lower limit. In other words, our integral should account for the region upto and including (0,0). But, in such case, the value of function on integral grows with no bound (tends to $\infty$). Hence, the finite value the question asks for can't be shown.

What am I missing here?

  • $\begingroup$ $1/\sqrt{x}$ grows to infinity as $x\to0$. Is $\int_0^1 dx/\sqrt{x}$ finite? $\endgroup$ Jul 22, 2017 at 2:43
  • $\begingroup$ Now, I am not sure. How did we ever get rid of the singularity at $x=0$ ? Can we even compute that limit since $1/\sqrt(x)$ is discontinuous at x = 0 ? $\endgroup$
    – Bibek_G
    Jul 22, 2017 at 2:51
  • 3
    $\begingroup$ The lack of a bounded integrand means it's an improper integral. To determine whether the integral exists (i.e., converges), change the outer lower limit from $0$ to $t$, where $t$ is unknown but positive, and take the limit of the integral, as $t \to 0^{+}$. $\endgroup$
    – quasi
    Jul 22, 2017 at 2:52
  • $\begingroup$ For example, by definition $$\int_0^1 dx/\sqrt{x}=\lim_{t \to o^{+}}\int_t^1 dx/\sqrt{x}=\lim_{t \to 0^{+}}2-2\sqrt{t}=2$$ $\endgroup$
    – quasi
    Jul 22, 2017 at 2:56
  • $\begingroup$ However for the integral you are working on, it doesn't converge, so the problem as stated is, in fact, wrong. Are you sure you copied it correctly? Is it from an actual published textbook, or just a class handout? $\endgroup$
    – quasi
    Jul 22, 2017 at 2:58

1 Answer 1


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{a}\int_{y}^{a}{1 \over x^{2} + y^{2}}\,\dd x\,\dd y = {\pi a \over 4}:\ {\large ?}}$.

\begin{align} \int_{0}^{a}\int_{y}^{a}{1 \over x^{2} + y^{2}}\,\dd x\,\dd y & = \int_{0}^{a}\int_{0}^{a}{\bracks{x > y} \over x^{2} + y^{2}}\,\dd x\,\dd y \\[5mm] & = {1 \over 2}\braces{% \int_{0}^{a}\int_{0}^{a}{\bracks{x > y} \over x^{2} + y^{2}}\,\dd x\,\dd y + \int_{0}^{a}\int_{0}^{a}{\bracks{y > x} \over y^{2} + x^{2}}\,\dd y\,\dd x} \\[5mm] & = {1 \over 2}\int_{0}^{a}\int_{0}^{a}{\dd x\,\dd y \over x^{2} + y^{2}} = {1 \over 2}\int_{0}^{a}{1 \over y}\int_{0}^{a/y}{\dd x \over x^{2} + 1}\,\dd y \\[5mm] & = {1 \over 2}\int_{0}^{a}{1 \over y}\,\arctan\pars{a \over y}\,\dd y\qquad \pars{~\mbox{It}\ \underline{diverges}\ 'logaritmically'~} \end{align}

  • 1
    $\begingroup$ Thanks. Also the comment on "dimensional analysis" was helpful. fwiw, a friend suggested, the integral was supposed to be $$ \int_0^a\int_y^a \frac{x}{x^2+y^2} dx dy = \frac{\pi a}{4} $$ $\endgroup$
    – Bibek_G
    Jul 22, 2017 at 3:37
  • $\begingroup$ @Bibek_G Thanks. That's correct. $\Huge\left\{\substack{\bullet\quad\bullet\\ \smile}\right\}$ $\endgroup$ Jul 22, 2017 at 4:39

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