Derivative of an integral function

1. Dec 4, 2012

muzialis

Hello there,

I have the function

$$f(t) = \int_0 ^{t} \frac{1}{\sqrt{t-\tau}} \mathrm{d}\tau$$

The integral can be found, yielding an expression as

$$f(t) = 2 \sqrt{t}$$

The question is, if I tried to calculate the derivative of $$f$$, using the "derivation under the integral " rule I get

$$\frac {\mathrm{d} f }{\mathrm{d}t} = \int_0 ^{t} \frac{\mathrm{d}}{\mathrm{d}t} \frac{1}{\sqrt{t-\tau}} \mathrm{d}\tau + \frac{1}{\sqrt{t-t}}$$

which is a problem I never run across.
I went through the derivation of the "derivation under the integral " but I am not getting anywhere, for the moment.
Anybody's help would be appreciated, thanks

2. Dec 4, 2012

Mute

Loosely speaking, the first term with the derivative inside the integral will generate a "$-1/\sqrt{t-t}$" term, as $d(t-\tau)^{-1/2}/dt = -d(t-\tau)^{-1/2}/d\tau$, so the diverging terms cancel out.

If you change variables to $y = t - \tau$ before taking the t derivative you also avoid the problem altogether.

3. Dec 4, 2012

muzialis

Mute,

as always, much appreciated.

thanks

4. Dec 4, 2012

lavinia

you can write out the Newton quotient in terms of t then write

(t + Δt-$\tau$)$^{-1/2}$ in a Taylor series

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