Where this $2\pi$ came from?

Question

In this link there are this equation (12.3.2): $B = \frac{\mu_0}{4\pi}\int_{wire}\frac{I\sin{\theta}dx}{r^2}$. And is said to substitute $r$ by $\sqrt{x^2+R^2}$ and $\sin{\theta}$ by $\frac{R}{\sqrt{x^2+R^2}}$.

This part is easy.

$B = \frac{\mu_0}{4\pi}\int_{wire}\frac{I\frac{R}{\sqrt{x^2+R^2}}dx}{(\sqrt{x^2+R^2})^2} \Rightarrow B = \frac{\mu_0}{4\pi}\int_{wire}\frac{I\frac{R}{\sqrt{x^2+R^2}}dx}{x^2+R^2} \Rightarrow B = \frac{\mu_0}{4\pi}\int_{wire}I\frac{Rdx}{\sqrt{x^2+R^2}(x^2+R^2)}$

$B = \frac{\mu_0}{4\pi}\int_{wire}I\frac{Rdx}{(x^2+R^2)^\frac{1}{2}(x^2+R^2)^1} \Rightarrow B = \frac{\mu_0}{4\pi}\int_{wire}I\frac{Rdx}{(x^2+R^2)^\frac{3}{2}}$

This last equation above is equal to the right side of the equation (12.3.5) in this same link, but for some reason $4\pi$ came to be $2\pi$, I sincerely don't know where this $2\pi$ came from, someone can explain me this?

Answer 1

When you have an even function $f$ over the real line, assuming (absolute) integrability, you have $$ \int_{-\infty}^\infty f(x)\;dx = 2\int_0^\infty f(x)\;dx $$ In your context, take $$ f(x)=\frac{C}{(x^2+R^2)^{3/2}},\quad C=\frac{\mu_0IR}{4\pi}\;. $$

Answer 2

Mathematically, the function $$ f(x) = \frac{R}{(x^2 + R^2)^{3/2}} $$ is symmetric around $x = 0,$ that is, if you look at the graph from $x=-\infty$ to $x=0$ it is the exact mirror image of the graph from $x=0$ to $x=\infty,$ and the areas under the two curves are equal: $$ \int_{-\infty}^0 \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} = \int_0^\infty \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}}.$$ This is something you can confirm with a simply change of variable from $x$ to $-x.$

But it is also supported by the physical intuition stated immediately after Equation $(12.3.2)$: the wire is symmetric around the point $O$ (and more importantly, though it is not said explicitly, the wire is symmetric around the line $OP$). Therefore any field that is contributed by the left half of the wire is just exactly the same as the field contributed by the right half.

To continue in excruciating detail:

$$ \begin{align} B &= \frac{\mu_0}{4\pi}\int_\text{wire}\frac{I\sin\theta\,\mathrm dx}{r^2} && \text{Equation $(12.3.2)$}\\ &= \frac{\mu_0 I}{4\pi}\int_\text{wire}\frac{\sin\theta\,\mathrm dx}{r^2} && \text{$I$ is a constant factor}\\ &= \frac{\mu_0 I}{4\pi}\int_\text{wire}\frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} && \sin\theta = \dfrac{R}{\sqrt{x^2 + R^2}}, \ r^2 = x^2 + R^2\\ &= \frac{\mu_0 I}{4\pi}\int_{-\infty}^\infty \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} && \text{Wire from $x=-\infty$ to $x=\infty$}\\ &= \frac{\mu_0 I}{4\pi}\left( \int_{-\infty}^0 \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} + \int_0^\infty \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} \right) && \\ &= \frac{\mu_0 I}{4\pi}\left( \int_0^\infty \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} + \int_0^\infty \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} \right) && \text{Symmetry of the integral}\\ &= \frac{\mu_0 I}{4\pi}\left(2 \int_0^\infty \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} \right) && \\ &= \frac{\mu_0 I}{2\pi} \int_0^\infty \frac{R\,\mathrm dx}{(x^2 + R^2)^{3/2}} && \text{Cancel a factor of $2$}\\ \end{align} $$

and voilá, you have Equation $(12.3.5).$