Calculus – Solving FoxTrot Bill Amend Problems

algebra-precalculuscalculusderivativesintegrationlimits

So I found this on the Wolfram website today:

cartoon

So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, $\LaTeX$ed:

$ \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} $
$ \displaystyle\sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1} $
$ \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u – 1 \right] \right] $
$ \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^4 \left( \rho \cos \phi \right) \rho^2 \sin \phi \, \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta$

Ideas

The degree of the numerator is $\frac{3}{2}$. The degree of the
denominator is $\frac{3}{2}$. The expression is of an indeterminate
form, namely $\frac{\infty}{\infty}$, so we use l'Hoptial's Rule.
No ideas, really.
Derivatives are always pretty easy, although this one is a bit
bashy. Basically bash Product and Chain Rules, etc.
Integration by Parts bash? The $\mathrm{d}\theta$ part is very
trivial. I mean for the $\rho$ and $\phi$.

Best Answer

I can derive the sum quite easily, using partial fractions and the residue theorem.

Note that $$\frac{k^2}{k^3+1} = \frac13 \left [\frac{1}{k+1} + \frac{2 k-1}{k^2-k+1}\right ]$$

Now, note that

$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1} = 1-\log{2}$$

For the other piece, note the fortuitous coincidence that

$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} = -\frac12 \sum_{k=-\infty}^{\infty} \frac{(-1)^{k} (2 k-1)}{k^2-k+1}$$

(One may see this by showing that the map $k \mapsto -k$ produces the $k+1$th term of the summand.)

Now, the sum on the RHS may be evaluated using residue theory. In general, one may show that

$$\sum_{k=-\infty}^{\infty} (-1)^k f(k) = -\pi \sum_n \operatorname*{Res}_{z=z_n} [\csc{\pi z} \, f(z)]$$

where the $z_n$ are the non-integral poles of $f$. In this case, the poles are at $z=e^{\pm i \pi/3}$, and we are also lucky to have an $f$ of the form $g'/g$, so that the residue at each pole is simply $1$. Therefore, we have

$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} = \frac{\pi}{2} \left ( \csc{\left (\pi e^{i \pi/3}\right )}+ \csc{\left (\pi e^{-i \pi/3}\right )} \right ) = \pi \, \text{sech}{\left ( \frac{\sqrt{3} \pi}{2}\right )}$$

Therefore,

$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} k^2}{k^3+1} = \frac13 \cdot \left [ 1-\log{2} + \pi \, \text{sech}{\left ( \frac{\sqrt{3} \pi}{2}\right )}\right ]$$

Related Solutions

[Math] Finding the volume inside an elyptical cylinder and a sphere

The equation for the ellipse states $x^2+y^2-2x=0$ (...) from which I have $\rho(\theta)= \displaystyle\frac{\cos\theta}{1+\cos^2\theta}$ with $0\leq\theta\leq\pi$.

The equation of the ellipse is \begin{equation*} 2x^{2}+y^{2}-2x=0 \end{equation*} from which I've obtained \begin{equation*} \rho (\theta)=\frac{2\cos \theta }{\cos ^{2}\theta +1},\qquad \text{with }-\pi/2 \leq \theta <\pi/2, \end{equation*}

because the tangent to the ellipse at $(x,y)=(0,0)$ is the vertical line $x=0$. With these corrections the volume integral becomes

\begin{eqnarray*} V &=&\int_{-\pi /2}^{\pi /2}\left( \int_{0}^{\frac{2\cos \theta }{\cos ^{2}\theta +1}}2\rho \sqrt{1-\rho ^{2}}d\rho \right) \,d\theta \\ &=&-\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left. \left( 1-\rho ^{2}\right) ^{3/2}\right\vert _{0}^{\frac{2\cos \theta }{\cos ^{2}\theta +1}}\,d\theta \\ &=&-\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( 1-\left( \frac{2\cos \theta }{ \cos ^{2}\theta +1}\right) ^{2}\right) ^{3/2}-1\,\ d\theta \end{eqnarray*} Simplifying, we have that

\begin{eqnarray*} V &=&\frac{2}{3}\pi -\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( \left( \frac{ \sin ^{2}\theta }{\cos ^{2}\theta +1}\right) ^{2}\right) ^{3/2}\,d\theta \\ &=&\frac{2}{3}\pi -\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( \frac{\sin ^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta \\ &=&\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin ^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta , \end{eqnarray*} because the integrand is an even function. The remaining integral can be converted into a rational function of $t=\tan \frac{\theta }{2}$ by the Weirstrass substitution $t=\tan \frac{ \theta }{2}$, which I then evaluated in SWP (Scientific Work Place):

\begin{eqnarray*} V &=&\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin ^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta ,\qquad t=\tan \frac{ \theta }{2} \\ &=&\frac{2}{3}\pi -\frac{64}{3}\int_{0}^{1}\frac{t^{6}}{\left( 1+t^{4}\right) ^{3}\left( 1+t^{2}\right) }\,dt \\ &=&\cdots \\ &=&\frac{4}{3}\pi -\frac{7}{12}\sqrt{2}\pi =\frac{16-7\sqrt{2}}{12}\pi . \end{eqnarray*}

ADDED. The integrand in $t$ can be expanded into partial fractions as follows

\begin{equation*} \frac{t^{6}}{\left( 1+t^{4}\right) ^{3}\left( 1+t^{2}\right) }=\frac{1}{8} \frac{t^{2}-1}{1+t^{4}}+\frac{1}{4}\frac{3+t^{2}}{\left( 1+t^{4}\right) ^{2}} -\frac{1}{2}\frac{1+t^{2}}{\left( 1+t^{4}\right) ^{3}}-\frac{1}{8\left( 1+t^{2}\right) }. \end{equation*}

ADDED 2. WolframAlpha computation of $V =\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta$ confirms the result above.

Calculus – Computing Complex Elliptic Integral

Suppose we have a finite continuous solenoid of radius $\rho$ and axial length $L$ where $I$ is the current carried by the wire and $n$ is the number of turns of wire per unit length. For convenience, we place the center of the solenoid at the origin and orient its axis to align with the $z$-axis.

The magnetic field $\vec{B}$ at the point $\vec{r}_{0}$ due to the solenoid is given by the iterated integral

$$\vec{B}{\left(\vec{r}_{0}\right)}=n\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\,\frac{\mu_{0}I}{4\pi}\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}},$$

where $\vec{r}$ is the position vector for a point with cylindrical coordinates $\left(\rho,\phi,z\right)$ on the surface of the solenoid:

$$\vec{r}=\rho\cos{\left(\phi\right)}\,\hat{x}+\rho\sin{\left(\phi\right)}\,\hat{y}+z\hat{z}.$$

Similarly, the position vector $\vec{r}_{0}$ of a point with cylindrical coordinates $\left(\rho_{0},\phi_{0},z_{0}\right)$ can be written as

$$\vec{r}_{0}=\rho_{0}\cos{\left(\phi_{0}\right)}\,\hat{x}+\rho_{0}\sin{\left(\phi_{0}\right)}\,\hat{y}+z_{0}\hat{z}.$$

A closed-form expression for $\vec{B}{\left(\vec{r}_{0}\right)}$ in terms of complete elliptic integrals is derived below.

For reference, the complete elliptic integrals are defined here as follows:

$$K{\left(\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}};~~~\small{\kappa\in\left(-1,1\right)},$$

$$E{\left(\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}};~~~\small{\kappa\in\left[-1,1\right]},$$

$$\Pi{\left(\nu,\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}};~~~\small{\kappa\in\left(-1,1\right)\land\nu\in\left(-\infty,1\right)}.$$

As you correctly stated above, it can be shown that the distance between the two points in cylindrical coordinates is given by

$$\|\vec{r}_{0}-\vec{r}\|=\sqrt{\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}}.$$

The expression you gave for the vector quantity $\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)$, however, is wrong, so let's go over its calculation in detail.

Recall the cylindrical unit vectors $\hat{\rho}$ and $\hat{\phi}$ may be given in terms of Cartesian unit vectors by

$$\hat{\rho}=\cos{\left(\phi\right)}\,\hat{x}+\sin{\left(\phi\right)}\,\hat{y},$$

$$\hat{\phi}=-\sin{\left(\phi\right)}\,\hat{x}+\cos{\left(\phi\right)}\,\hat{y}.$$

Then, $\vec{r}=\rho\hat{\rho}+z\hat{z}$ and $\frac{\partial\vec{r}}{\partial\phi}=\rho\hat{\phi}$.

Similarly, $\vec{r}_{0}=\rho_{0}\hat{\rho}_{0}+z_{0}\hat{z}$, where the unit vectors $\hat{\rho}_{0}$ and $\hat{\phi}_{0}$ can be expressed as

$$\hat{\rho}_{0}=\cos{\left(\phi_{0}\right)}\,\hat{x}+\sin{\left(\phi_{0}\right)}\,\hat{y},$$

$$\hat{\phi}_{0}=-\sin{\left(\phi_{0}\right)}\,\hat{x}+\cos{\left(\phi_{0}\right)}\,\hat{y}.$$

The above pair of equations can be inverted to give $\hat{x}$ and $\hat{y}$ in terms of $\hat{\rho}_{0}$ and $\hat{\phi}_{0}$:

$$\hat{x}=\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0},$$

$$\hat{y}=\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}.$$

We then find

$$\begin{align} \hat{\rho} &=\cos{\left(\phi\right)}\,\hat{x}+\sin{\left(\phi\right)}\,\hat{y}\\ &=\cos{\left(\phi\right)}\left[\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]+\sin{\left(\phi\right)}\left[\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]\\ &=\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0},\\ \end{align}$$

$$\begin{align} \hat{\phi} &=-\sin{\left(\phi\right)}\,\hat{x}+\cos{\left(\phi\right)}\,\hat{y}\\ &=-\sin{\left(\phi\right)}\left[\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]+\cos{\left(\phi\right)}\left[\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]\\ &=\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}.\\ \end{align}$$

and thus,

$$\begin{align} \hat{\phi}\times\left(\vec{r}_{0}-\vec{r}\right) &=\hat{\phi}\times\left(\rho_{0}\hat{\rho}_{0}+z_{0}\hat{z}-\rho\hat{\rho}-z\hat{z}\right)\\ &=\hat{\phi}\times\left[\left(z_{0}-z\right)\hat{z}-\rho\hat{\rho}+\rho_{0}\hat{\rho}_{0}\right]\\ &=\left(z_{0}-z\right)\hat{\phi}\times\hat{z}-\rho\hat{\phi}\times\hat{\rho}+\rho_{0}\hat{\phi}\times\hat{\rho}_{0}\\ &=\left(z_{0}-z\right)\hat{\rho}+\rho\hat{z}-\rho_{0}\hat{\rho}_{0}\times\hat{\phi}\\ &=\left(z_{0}-z\right)\left[\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\right]+\rho\hat{z}\\ &~~~~~-\rho_{0}\hat{\rho}_{0}\times\left[\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\right]\\ &=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\rho\hat{z}\\ &~~~~~-\rho_{0}\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}\times\hat{\rho}_{0}-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}\times\hat{\phi}_{0}\\ &=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\rho\hat{z}\\ &~~~~~-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\,\hat{z}\\ &=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\\ &~~~~~+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}.\\ \end{align}$$

We turn now to the calculation of $\vec{B}{\left(\vec{r}_{0}\right)}$ for the solenoid, and we start by simplifying the angular integral.

Given $\vec{r}_{0}\in\mathbb{R}^{3}$ such that $\neg\left[\rho_{0}=\rho\land\left(|z_{0}|\le\frac{L}{2}\right)\right]$, we have

$$\begin{align} \vec{B}{\left(\vec{r}_{0}\right)} &=n\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\,\frac{\mu_{0}I}{4\pi}\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho\hat{\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}\bigg{]}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}\bigg{]}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]},\\ \end{align}$$

where in the second-to-last line above we used the fact that the $\hat{\phi}_{0}$ component of the angular integral is identically zero because of its $2\pi$-periodic antiderivative, and in the last line above we've used the following lemma: for any $a\in\mathbb{R}$ and any continuous periodic function $f:\mathbb{R}\rightarrow\mathbb{R}$ with period $p\in\mathbb{R}_{>0}$, it can be shown that

$$\int_{0}^{p}\mathrm{d}x\,f{\left(x+a\right)}=\int_{0}^{p}\mathrm{d}x\,f{\left(x\right)}.$$

Then,

$$\begin{align} \vec{B}{\left(\vec{r}_{0}\right)} &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{z_{0}-\frac{L}{2}}^{z_{0}+\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]};~~~\small{\left[z\mapsto z_{0}-z\right]}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{z_{0}-\frac{L}{2}}^{z_{0}+\frac{L}{2}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]};~~~\small{symmetry}\\ &=B_{0}\int_{z_{-}}^{z_{+}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}},\\ \end{align}$$

where in the last line above we've set $B_{0}:=\frac{\mu_{0}nI}{4\pi}$ and $z_{\pm}:=z_{0}\pm\frac{L}{2}$.

Now, we could proceed to evaluate the angular integral in terms of elliptic integrals as others have already pointed out, but that would leave us with a very difficult non-elementary axial integral. It turns out, however, that changing the order of integration makes the integral much easier.

Consider the following derivative:

$$\frac{d}{dz}\left[\frac{z\vec{c}-a\vec{b}}{a\sqrt{z^{2}+a}}\right]=\frac{z\vec{b}+\vec{c}}{\left(z^{2}+a\right)^{3/2}};~~~\small{a\in\mathbb{R}_{>0}\land\vec{b}\in\mathbb{R}^{3}\land\vec{c}\in\mathbb{R}^{3}}.$$

Assuming $0<\rho_{0}\neq\rho$, we have $0<\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}$. Also note that $0<\frac{4\rho_{0}\rho}{z^{2}+\left(\rho_{0}+\rho\right)^{2}}<1$ for all real $z$.

Setting $\nu:=\frac{4\rho_{0}\rho}{\left(\rho_{0}+\rho\right)^{2}}\land k_{\pm}:=\sqrt{\frac{4\rho_{0}\rho}{z_{\pm}^{2}+\left(\rho_{0}+\rho\right)^{2}}}$, we then have

$$\begin{align} \vec{B}{\left(\vec{r}_{0}\right)} &=B_{0}\int_{z_{-}}^{z_{+}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\int_{z_{-}}^{z_{+}}\mathrm{d}z\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\int_{z_{-}}^{z_{+}}\mathrm{d}z\,\frac{d}{dz}\bigg{[}-\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\ &~~~~~+\frac{2z\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\bigg{]}\\ &=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\,\bigg{[}-\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}+\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\ &~~~~~+\frac{2z_{+}\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\ &~~~~~-\frac{2z_{-}\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\bigg{]}\\ &=2B_{0}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}-\frac{2\rho\cos{\left(2\varphi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\ &~~~~~+\frac{2\rho\cos{\left(2\varphi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\ &~~~~~+\frac{2z_{+}\rho\left[\rho-\rho_{0}\cos{\left(2\varphi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}\right]\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\ &~~~~~-\frac{2z_{-}\rho\left[\rho-\rho_{0}\cos{\left(2\varphi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}\right]\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\bigg{]};~~~\small{\left[\phi=2\varphi\right]}\\ &=2B_{0}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{\left[2\rho-4\rho\cos^{2}{\left(\varphi\right)}\right]\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{\left[2\rho-4\rho\cos^{2}{\left(\varphi\right)}\right]\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~+\frac{\left[-\rho_{0}^{2}+\rho^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]z_{+}\hat{z}}{\left[\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]\sqrt{z_{+}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{\left[-\rho_{0}^{2}+\rho^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]z_{-}\hat{z}}{\left[\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]\sqrt{z_{-}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\bigg{]}\\ &=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{2\rho\left[1-2\cos^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\cos^{2}{\left(\varphi\right)}}}-\frac{2\rho\left[1-2\cos^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~+\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\cos^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\cos^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\cos^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\cos^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\cos^{2}{\left(\varphi\right)}}}\bigg{]},\\ \end{align}$$

and finally,

$$\begin{align} \vec{B}{\left(\vec{r}_{0}\right)} &=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}-\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~+\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\bigg{]};~~~\small{\left[\varphi\mapsto\frac{\pi}{2}-\varphi\right]}\\ &=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~+\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &=\hat{\rho}_{0}\frac{4B_{0}\sqrt{\rho}}{k_{+}\sqrt{\rho_{0}}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}\right]-\left(1-\frac{k_{+}^{2}}{2}\right)}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\hat{\rho}_{0}\frac{4B_{0}\sqrt{\rho}}{k_{-}\sqrt{\rho_{0}}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}\right]-\left(1-\frac{k_{-}^{2}}{2}\right)}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~+\hat{z}\frac{B_{0}k_{+}z_{+}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\hat{z}\frac{B_{0}k_{-}z_{-}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &=\frac{4B_{0}\sqrt{\rho}}{k_{+}\sqrt{\rho_{0}}}\bigg{[}E{\left(k_{+}\right)}-\left(1-\frac{k_{+}^{2}}{2}\right)K{\left(k_{+}\right)}\bigg{]}\hat{\rho}_{0}\\ &~~~~~-\frac{4B_{0}\sqrt{\rho}}{k_{-}\sqrt{\rho_{0}}}\bigg{[}E{\left(k_{-}\right)}-\left(1-\frac{k_{-}^{2}}{2}\right)K{\left(k_{-}\right)}\bigg{]}\hat{\rho}_{0}\\ &~~~~~+\frac{B_{0}k_{+}z_{+}}{\sqrt{\rho_{0}\rho}}\bigg{[}K{\left(k_{+}\right)}-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)\Pi{\left(\nu,k_{+}\right)}\bigg{]}\hat{z}\\ &~~~~~-\frac{B_{0}k_{-}z_{-}}{\sqrt{\rho_{0}\rho}}\bigg{[}K{\left(k_{-}\right)}-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)\Pi{\left(\nu,k_{-}\right)}\bigg{]}\hat{z}.\blacksquare\\ \end{align}$$

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