Simplifying the Integral of d lnp and Understanding its Truth

[sparkle123]

I don't understand why this is true:

My answer:
integral(d lnp) from lnp* to lnp
=lnp from lnp* to lnp
=ln(lnp)-ln(lnp*)

 

[Dick]

I think you are right.

 

[sparkle123]

it's from a textbook though :(

 

[Dick]

it's from a textbook though :(

So? integral d(f(x))=f(x)+C, right? People write textbooks. People make mistakes.

 

[I like Serena]

I don't understand why this is true:
My answer:
integral(d lnp) from lnp* to lnp
=lnp from lnp* to lnp
=ln(lnp)-ln(lnp*)

Substitute ln p = u and ln p* = u*
Then we have:

\int_{\ln p^*}^{\ln p} d \ln p = \int_{u^*}^u du = u - u^* = \ln p - \ln p^* = \ln \frac p {p^*}

[EDIT]or written in another way:

\int_{\ln p^*}^{\ln p} d \ln p = \int_{p^*}^p \frac 1 p dp = \ln p - \ln p^* = \ln \frac p {p^*}

[/EDIT]

 

[sparkle123]

Thanks!

 

[Dick]

Substitute ln p = u and ln p* = u*
Then we have:

\int_{\ln p^*}^{\ln p} d \ln p = \int_{u^*}^u du = u - u^* = \ln p - \ln p^* = \ln \frac p {p^*}

[EDIT]or written in another way:

\int_{\ln p^*}^{\ln p} d \ln p = \int_{p^*}^p \frac 1 p dp = \ln p - \ln p^* = \ln \frac p {p^*}

[/EDIT]

Ok, so you should take the limits to be the limits of ln(p). Not the limits of p. Sloppy on my part. Thanks for the correction.