Integral with an inverse function limit

  • #1
18
0
Hello,

I have tried the integral below with Mathematica and it gives me the following solution:

##\frac{d}{dc}\int_{z^{-1}(c)}^{1} z(x)dx = -\frac{c}{z'(z^{-1}(c))}##

I am not quite sure where it gets it from...I think it can be separated and with differentiation the first part will be zero:

##\frac{d}{dc}\int_{z^{-1}(c)}^{1}z(x)dx=\frac{d}{dc}\int_{0}^{1} z(x)dx-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx=-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx##

By the formula ##\frac{d}{dc}z^{-1}(c) = \frac{1}{z'(z^{-1}(c))}##, so it definitely has to play a role here, but how? and where does c come from? Do we need to somehow use chain rule here?

If the upper limit was just c it would be easy to take it with a Leibniz rule:

##-\frac{d}{dc}\int_{0}^{c} z(x)dx=z(c)##

However, it seems that when one of the limits is an inverse function, there is something different at work which I am not aware of...
 

Answers and Replies

  • #2
193
29
Let me struggle through this in my own way.
I find it useful to give familiar-looking names to things wherever possible.
So let's put
z-1(c) = y
and ∫z(x)dx = Z,
and I'll call the integral-derivative we have to evaluate F.

So we have to evaluate F = (d/dc)( Z(1) -Z(y) ),
which immediately becomes
-(d/dc)Z(y) = -(dZ(y)/dy). (dy/dc)
= z(y).(dy/dc)

But z(y) = z(z-1(c)) = c
So we have
F = c.(dy/dc) . . . . . . .(A)

Now z-1(c) = y by definition; so
c = z(y), and, differentiating,
dc/dy =dz/dy,
or if we invert,
dy/dc = 1/(dz/dy) = 1/z'(y)
= 1/z'( z-1(c) )

Substitute this for dy/dc in (A) and I think we're done.
 
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  • #3
18
0
Thank you, it makes sense now!
 

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