Understanding this Lagrangian calculation I was trying to understand this section of a Wikipedia article:

$$0 = \delta \int \sqrt{2T} d\tau =
  \int \frac{\delta T}{\sqrt{2T}} d\tau =
  \frac{1}{c} \delta \int T d\tau$$

For the life of me, I can't figure out how does one get from $\displaystyle \delta \int \sqrt{2T} d\tau$ to $\displaystyle \int \frac{\delta T}{\sqrt{2T}} d\tau$. What is $\delta$? I assume it is some kind of differential operator but in respect to what variable?
 A: 
I was trying to understand this section of a Wikipedia article:


$$0 = \delta \int \sqrt{2T} d\tau =
  \int \frac{\delta T}{\sqrt{2T}} d\tau =
  \frac{1}{c} \delta \int T d\tau$$


For the life of me, I can't figure out how does one get from $\displaystyle \delta \int \sqrt{2T} d\tau$ to $\displaystyle \int \frac{\delta T}{\sqrt{2T}} d\tau$.

Let the function $T(\tau)$ be changed to another function that is "close" to $T(\tau)$:
$$
T(\tau) \to T(\tau) + \epsilon \eta(\tau)\;,
$$
where $\eta(\tau)$ is a function and $\epsilon$ is "small" constant in a sense we will describe more below.
We note that the relationship between our notation and the  Wikipedia notation is:
$$
\delta T = \epsilon \eta(\tau)\;,
$$
where our notation makes explicit that the variation $\delta T$ of $T$ is an arbitrary function  $\eta(\tau)$ times a "small" expansion parameter $\epsilon$.
We literally make the above replacement in the integral of interest:
$$
I = \int \sqrt{2T}d\tau \to \int \sqrt{2T+2\epsilon \eta(\tau)}d\tau\;,
$$
and then we look at only the first order change in the integral with respect to $\epsilon$. We look at the first order change because the integrand of the first order change in $I$ is what we define as the functional derivative of $I$ (in analogy with the derivative of a multivariate function). We are "allowed" to do this because $\epsilon$ is "small" in the sense that we can ignore all the higher order terms in the $\epsilon$ expansion.
Expanding to first order in epsilon can be performed in this case via the known McLauren/Taylor series for the function $\sqrt{1+x}$. (Which is $\sqrt{1+x} \approx 1 + x/2 + \ldots$, where the terms being ignored are of order $x^2$ and higher.)
$$
\int \sqrt{2T+2\epsilon \eta(\tau)}d\tau 
=\int \sqrt{2T}\sqrt{1+\frac{\epsilon \eta(\tau)}{T}}d\tau 
\approx \int \sqrt{2T}\left({1+\frac{\epsilon \eta(\tau)}{2T}}\right)d\tau
\equiv I + \delta I\;,
$$
where, now we see that:
$$
\delta I = \int \sqrt{2T}\left({\frac{\epsilon \eta(\tau)}{2T}}\right)d\tau
=\int \frac{\epsilon \eta(\tau)}{\sqrt{2T}}d\tau
\equiv\int \frac{\delta T}{\sqrt{2T}}d\tau\;.
$$
We could also write this result as a functional derivative, by definition of the latter, as:
$$
\frac{\delta I}{\delta T(\tau)} = \frac{1}{\sqrt{2T}}
$$
A: $T(\tau)$ is a function. $dT(\tau)$ is also a function that is "small" i.e. less than some $\epsilon$ for all $\tau$. $T(\tau) + \delta T(\tau)$ means you perturb the function by adding a small perturbation at all $\tau$. If you do this then $\sqrt{2(T + \delta T)} \approx \frac{\delta T}{\sqrt{2T}}$. Here you kind of Taylor expand the function at each $\tau$ (also the arguments of the functions are suppressed for brevity). $\delta(F(T))$ means the expression you get when substituting $T + \delta T$ for $T$ in $F$, so $\delta\sqrt{2T} \equiv \sqrt{2(T+\delta T)}$.
To better understand all of this I would suggest reading a good textbook on the calculus of variations. There are many around (this is a relatively old discipline of mathematics).
