Derivative of an integral function

[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

 

[Mute]

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

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.

 

[muzialis]

Mute,

as always, much appreciated.

thanks

 

[lavinia]

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

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