Courant John II 4.6 Problem 16.

Prove that $$\iint_R e^{-(x^2+y^2)}\,dx\, dy=ae^{-a^2}\int_0^{\infty}\frac{e^{-u^2}}{a^2+u^2}\,du\,,$$ where $R$ denotes the half-plane $x\ge a\gt 0$, by applying the transformation $x^2+y^2=u^2+a^2, y=vx$.

I did the transformation to get $$\iint_R e^{-(x^2+y^2)}\,dx \,dy=2\int_0^{\infty}\int_0^{u/a}ue^{-(u^2+a^2)}\left( 1+\dfrac1{v^2} \right)\,dv \,du,$$ but the inner integral (omitting constant part independent to v) $$\int_0^{u/a}\left( 1+\dfrac1{v^2} \right)dv$$ is a divergent improper integral. I have no idea what to now. I also tried rewriting so the inner integral is with respect to u instead, but the exact same problem still occurs.

(EDIT): The Jacobian $$J=x_uy_v-x_vy_u=\left(\dfrac ux \right)(x)-\left(-\dfrac{y}{v^2} \right)\left(\dfrac uy \right)=u \left( 1+\dfrac1{v^2} \right)$$ since $x=\sqrt{u^2+a^2-y^2}$

(EDIT) Correction of Jacobian: $J=\dfrac{u}{1+v^2}$

  • $\begingroup$ How did you get the Jacobian? I'm not sure, but I think that might be the issue. $\endgroup$ May 14, 2020 at 15:10
  • $\begingroup$ Just edited post to show it. $\endgroup$
    – Divide1918
    May 14, 2020 at 15:18
  • $\begingroup$ That's the issue. You should express $x$ in terms of just $u$ and $v$. The same with $y$. For instance you are saying $y=vx$ and then $y_v=x$. This is wrong, since $x$ also depends on $v$. By product rule $y=vx$ implies $y_v=x+vx_v$. $\endgroup$ May 14, 2020 at 15:25
  • $\begingroup$ @Divide1918 I took the liberty of improving your title and formatting. Descriptive titles are better and help with searches. Hope you find your answer. $\endgroup$ May 14, 2020 at 17:11

2 Answers 2


It looks like you can just integrate by parts. Your integral (with the correct Jacobian) is $$2e^{-a^2} \int_0^\infty u e^{-u^2} \left( \int_0^{u/a} \frac{1}{1+v^2}\, dv \right) \, du.$$

Let $f'(u) = ue^{-u^2}$ and $g(u) = \displaystyle \int_0^{u/a} \frac{1}{1+v^2} \, dv$. Then (modulu constants) $f(u) = - \frac 12 e^{-u^2}$ and $$g'(u) = \frac{1}{a} \frac{1}{1 + (u/a)^2} = \frac{a}{a^2 + u^2}.$$ Since $f(0) g(0) = 0$ and $\lim_{u \to \infty} f(u) g(u) = 0$ ($g$ is bounded) you get $$\int_0^\infty f'(u) g(u) \, du = \frac a2 \int_0^\infty \frac{e^{-u^2}}{a^2 + u^2} \, du.$$


The correct Jacobian is $J=\frac{u}{1+v^2}$. So, your integral becomes $$I=2\int_0^\infty \int_0^{u/a}\frac{u}{1+v^2}e^{-(u^2+a^2)}dvdu$$ Let's change the order of the integrals, note that $\{0\leq u< \infty$, $0\leq v\leq u/a\}$ is the same as $\{0\leq v<\infty$, $av\leq u<\infty\}$ So, your integral becomes

$$I=2\int_0^\infty \int_{av}^\infty\frac{u}{1+v^2}e^{-(u^2+a^2)}dudv$$ We can evaluate this integral (note that $\int ue^{-(u^2+a^2)}du=\frac{-1}{2}e^{-(u^2+a^2)}+c$) \begin{align}I&=2\int_0^\infty \frac{1}{1+v^2}\frac{-1}{2}e^{-(u^2+a^2)}\Big|^\infty_{av}dv\\ &=\int_0^\infty \frac{e^{-a^2(v^2+1)}}{v^2+1}dv\\ &=e^{-a^2}\int_0^\infty \frac{e^{-a^2(v^2)}}{v^2+1}dv \end{align} After the change of variable $v=w/a$, you get \begin{align}I&=e^{-a^2}\int_0^\infty \frac{e^{-w^2}}{(w/a)^2+1}\frac{dw}{a}\\ &=ae^{-a^2}\int_0^\infty \frac{e^{-w^2}}{w^2+a^2}dw \end{align} Which is what we wanted to prove.

About the correct Jacobian Here I show how to compute the correct Jacobian without too many computations. From $y=vx$, by taking partial derivatives, we get $y_v=x+vx_v$ and $y_u=vx_u$. Hence, $$J=x_uy_v-x_vy_u=x_u(x+vx_v)-x_v(vx_u)=xx_u.$$ Now, on the other hand, since $x^2+y^2=u^2+a^2$, we have $x^2+(vx)^2=u^2+a^2$, i.e. $(1+v^2)x^2=u^2+a^2$. Taking partial derivative with respect to $u$ in both sides, we get $(1+v^2)2x x_u=2u.$ So, we conclude that $$J=xx_u=\frac{u}{1+v^2}.$$


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