$\int_{-\infty}^\infty \frac{\lambda l}{2\pi\epsilon_0(x^2+l^2)^{3/2}}dx$ Proving an electric field from a wire falls off at $1/l$ If you don't know the physics behind all this, that's okay, I just need the integral of this function (or limit, I'm not too sure).
Here's the gist: normally with infinitesimal point charges, there is an electric field (represented by vectors everywhere in space) that points radially outward from the position of the point, and its magnitude falls off as $\frac{1}{r^2}$. Here is the whole equation for you just to give you some background:
$$E = \frac{q}{4*\pi*\epsilon_0*r^2}$$
Where $q$ and $\epsilon_0$ are just constants.
Now imagine a continuous straight line of these infinitesimally small particles, giving it one dimension. I am trying to find the electric field at one point as a product of all these particles integrated together (from infinity to negative infinity).
As my variable in the integration, ill use $x$ - the length of the wire traversed so far. And to replace $q$, ill use $\lambda*dx$, making $\lambda$ a constant density of charge per unit length (simulating the point charges as $dx$ approaches 0).
Now this means that $r$ will turn into more of a Pythagorean theorem, as we are trying to find the electric field at one point caused from the whole wire. To replace $r$, ill use $l$ and $x$ ($l$ being the perpendicular length away from the wire). Thus $r$ then becomes $\sqrt{x^2+l^2}$.
Now here's where it gets a little tricky. The electric field calculated from each point is a vector, and points in a direction. But because we have a one dimensional wire as the source of each field, that means that the E-field caused from $x$ will cancel out the field caused from
$-x$ (but it won't cancel out the e-field in the direction perpindicular to the wire). For this we just multiply the magnitude of the electric field by $cos(\theta)$, with $\theta$ being the angle difference between the normal vector (90 degrees, perpindicular to the wire) and the electric field vector.
Now an even more fun problem! How do we get $\theta$? Through the inverse tangent of $x$ and $l$! And what do we end up getting?
$$\cos(\theta) = \cos(\arctan(x/l)) = \frac{1}{\sqrt{1+(x/l)^2}}$$
But the thing is that there are two sides that cancel out, doubling the value we get from the cosine.
So now let's recap. Here is the revised equation so far:
$$E = \frac{\lambda*dx}{2*\pi*\epsilon_0*\sqrt{x^2+l^2}^2*\sqrt{1+(x/l)^2}}$$
$\sqrt{x^2+l^2}^2$ simplifies to $(x^2+l^2)$, and $\sqrt{1+(x/l)^2}$ simplifies to $\frac{\sqrt{l^2+x^2}}{l}$. Thus, the whole equation is now
$$E = \frac{\lambda*dx*l}{2*\pi*\epsilon_0*(x^2+l^2)^{3/2}}$$
Where $\lambda$ and $\epsilon_0$ are constants.
Now my question is, "what is the integration of this function over infinity, and how do I do it?".
$$\int_{-\infty}^\infty \frac{\lambda\cdot dx\cdot l}{2\pi\epsilon_0(x^2+l^2)^{3/2}}$$
I really appreciate your help or any tips! Thank you!!!
 A: Great set up.  You can probably integrate with trig substitution but I'd use Wolfram.  On the other hand, Gauss's law will give you the answer with almost no work:
$$\oint{E\cdot da} = \frac{1}{\epsilon}Q_{enc}$$ where $Q_{enc}$ is the charge enclosed.   For an infinite wire, we surround it by an infinite cylinder.  Suppose for the moment the cylinder has height h (this will drop out anyway) and radius r (this is how far away the point of interest is).  Then, since the charge density is uniform, E is uniform across the surface of the cylinder.  Ignoring the top and bottom (E is perpendicular there and so they drop out of the dot product in the integral) we have: $$|E|\oint da = \frac{1}{\epsilon} \lambda h$$
$$ |E| 2 \pi r h = \frac{1}{\epsilon} \lambda h$$ and so $$|E| = \frac{\lambda}{2\pi\epsilon r}$$ A simple symmetry argument will tell you the direction of the field is entirely perpendicular to the wire.
A: Evaluation of the integral
$$
E=\frac{1}{2\pi \epsilon _{0}}\lambda l\int_{-\infty }^{\infty }\frac{dx}{
(x^{2}+l^{2})^{3/2}}.\tag{1}
$$
Use the substitution $x=l\tan t$. Then
$$
\frac{dx}{dt}=l\left( 1+\tan ^{2}t\right) =l\sec ^{2}t=\frac{l}{\cos ^{2}t}
$$
and 
$$
\begin{eqnarray*}
I &=&\int \frac{dx}{(x^{2}+l^{2})^{3/2}} \\
&=&\int \frac{1}{(l^{2}\tan ^{2}t+l^{2})^{3/2}}l\left( 1+\tan ^{2}t\right)
\,dt \\
&=&\int \frac{1}{l^{2}\sqrt{1+\tan ^{2}t}}\,dt=\int \frac{\cos t}{l^{2}}\,dt=
\frac{1}{l^{2}}\sin t+C \\
&=&\frac{x}{l^{2}\sqrt{x^{2}+l^{2}}}+C,\tag{2}
\end{eqnarray*}
$$
because
$$
\sin t=\frac{\tan t}{\sec t}=\frac{\tan t}{\sqrt{1+\tan ^{2}t}}=\frac{x/l}{\sqrt{1+x^{2}/l^{2}}}=\frac{x}{\sqrt{x^{2}+l^{2}}}.
$$
So 
$$
\begin{eqnarray*}
J &=&\int_{-\infty }^{\infty }\frac{1}{(x^{2}+l^{2})^{3/2}}
dx=2\int_{0}^{\infty }\frac{1}{(x^{2}+l^{2})^{3/2}}dx \\
&=&\left. \frac{2x}{l^{2}\sqrt{x^{2}+l^{2}}}\right\vert _{0}^{\infty
}=2\lim_{x\rightarrow \infty }\frac{x}{l^{2}\sqrt{x^{2}+l^{2}}}-0=\frac{2}{l^{2}},\tag{3}
\end{eqnarray*}
$$
because the integrand is an even function. The integral $(1)$ is thus
$$
E=\frac{1}{2\pi \epsilon _{0}}\lambda l\times J=\frac{\lambda }{\pi \epsilon
_{0}}\frac{1}{l}.\tag{4}
$$
Non mathematical remark: In comparison with Ethan's answer the difference of a factor of $1/2$ is due to the stated form for the integral $E$. You have
doubled the integrand: "the thing is that there are two sides that cancel
out, doubling the value we get from the cosine", but at the same time
integrated from $-\infty $ to $+\infty $. In my opinion we should compute
the magnitude of the electric field vector as $$\frac{1}{4\pi \epsilon _{0}}\lambda l\int_{-\infty }^{\infty }\frac{dx}{(x^{2}+l^{2})^{3/2}}\qquad\text{or}\qquad\frac{1
}{2\pi \epsilon _{0}}\lambda l\int_{0}^{\infty }\frac{dx}{(x^{2}+l^{2})^{3/2}
}.$$ See, e.g. Electric Field Near an Infinitely Long Line Charge.
