Derivative inside an integral problem. Need help.

[Diogenes]

Could someone please tell me where I am going wrong here?
I am trying to take the derivative with respect to \(\displaystyle T\) of this equation:
\(\displaystyle \begin{align*} U = \sum_p \int \displaystyle \frac{\hbar \omega}{e^{\frac{\hbar \omega}{k_b T}}-1} D_p(\omega) dw\end{align*}\)


By making a change of variable \(\displaystyle x=\frac{\hbar \omega}{k_b T}\)
And the book shows the result to be


\(\displaystyle \begin{align*}\frac{dU}{dT} = k_b \sum_p \int \frac{x^2 e^x}{(e^x-1)^2} D_p(\omega) dw\end{align*}\)

So the way I went about this was to first make the change of variable substitution for \(\displaystyle x\) giving

\(\displaystyle \begin{align*}U = \sum_p \int \frac{x k_b T}{e^x-1} D_p(\omega) dw\end{align*}\)

and then use the chain rule inside the integral

\(\displaystyle \begin{align*}\frac{dU}{dT} & = \frac{dx}{dT} \frac{dU}{dx} \\&= -\frac{\hbar \omega}{k_b T^2} \frac{dU}{dT}\\&= -\frac{\hbar \omega}{k_b T^2} \frac{d}{dT} (\sum_p \int \frac{x k_b T}{e^x-1} D_p(\omega) dw )\\&= -\frac{\hbar \omega}{k_b T^2} \sum_p \int \frac{d}{dT} ( \frac{x k_b T}{e^x-1} D_p(\omega) dw )\\&= -\frac{\hbar \omega}{k_b T^2} \sum_p \int \frac{k_b T(e^x-1)-e^x x k_b T}{(e^x-1)^2} D_p(\omega) dw \\&= -\frac{\hbar \omega}{k_b T^2} - k_b T \sum_p \int \frac{e^x x - e^x +1}{(e^x-1)^2} D_p(\omega) dw \\&= \frac{\hbar \omega}{k_b T} k_b \sum_p \int \frac{e^x x - e^x +1}{(e^x-1)^2} D_p(\omega) dw \\&= k_b x \sum_p \int \frac{e^x x - e^x +1}{(e^x-1)^2} D_p(\omega) dw \\&= k_b \sum_p \int \frac{e^x x^2 - e^x x +x}{(e^x-1)^2} D_p(\omega) dw \\\end{align*}\)

Which obviously doesn't equal this
\(\displaystyle \begin{align*}\frac{dU}{dT} = k_b \sum_p \int \frac{x^2 e^x}{(e^x-1)^2} D_p(\omega) dw\end{align*}\)

Really at a loss here can't find my mistake. Thanks in advance.

 

[HallsofIvy]

Could someone please tell me where I am going wrong here?
I am trying to take the derivative with respect to \(\displaystyle T\) of this equation:
\(\displaystyle \begin{align*} U = \sum_p \int \displaystyle \frac{\hbar \omega}{e^{\frac{\hbar \omega}{k_b T}}-1} D_p(\omega) dw\end{align*}\)


By making a change of variable \(\displaystyle x=\frac{\hbar \omega}{k_b T}\)
And the book shows the result to be


\(\displaystyle \begin{align*}\frac{dU}{dT} = k_b \sum_p \int \frac{x^2 e^x}{(e^x-1)^2} D_p(\omega) dw\end{align*}\)

So the way I went about this was to first make the change of variable substitution for \(\displaystyle x\) giving

\(\displaystyle \begin{align*}U = \sum_p \int \frac{x k_b T}{e^x-1} D_p(\omega) dw\end{align*}\)

That makes no sense. Your "substitution" replaces "T" by "x" so you should NOT have a "T" in the result.
How did "\(\displaystyle \hbar \omega\)", a constant times the dummy variable, come to depend on x and T?

and then use the chain rule inside the integral

\(\displaystyle \begin{align*}\frac{dU}{dT} & = \frac{dx}{dT} \frac{dU}{dx} \\&= -\frac{\hbar \omega}{k_b T^2} \frac{dU}{dT}\\&= -\frac{\hbar \omega}{k_b T^2} \frac{d}{dT} (\sum_p \int \frac{x k_b T}{e^x-1} D_p(\omega) dw )\\&= -\frac{\hbar \omega}{k_b T^2} \sum_p \int \frac{d}{dT} ( \frac{x k_b T}{e^x-1} D_p(\omega) dw )\\&= -\frac{\hbar \omega}{k_b T^2} \sum_p \int \frac{k_b T(e^x-1)-e^x x k_b T}{(e^x-1)^2} D_p(\omega) dw \\&= -\frac{\hbar \omega}{k_b T^2} - k_b T \sum_p \int \frac{e^x x - e^x +1}{(e^x-1)^2} D_p(\omega) dw \\&= \frac{\hbar \omega}{k_b T} k_b \sum_p \int \frac{e^x x - e^x +1}{(e^x-1)^2} D_p(\omega) dw \\&= k_b x \sum_p \int \frac{e^x x - e^x +1}{(e^x-1)^2} D_p(\omega) dw \\&= k_b \sum_p \int \frac{e^x x^2 - e^x x +x}{(e^x-1)^2} D_p(\omega) dw \\\end{align*}\)

Which obviously doesn't equal this
\(\displaystyle \begin{align*}\frac{dU}{dT} = k_b \sum_p \int \frac{x^2 e^x}{(e^x-1)^2} D_p(\omega) dw\end{align*}\)

Really at a loss here can't find my mistake. Thanks in advance.

 

[Diogenes]

That makes no sense. Your "substitution" replaces "T" by "x" so you should NOT have a "T" in the result.
How did "\(\displaystyle \hbar \omega\)", a constant times the dummy variable, come to depend on x and T?

By the change of variable I suppose. Since we are calling \(\displaystyle \displaystyle x=\frac{\hbar \omega}{k_b T}\), then solving for \(\displaystyle \hbar \omega\) we have \(\displaystyle x k_b T=\hbar \omega\). Then replace that in the numerator of the integrand of \(\displaystyle \displaystyle U = \sum_p \int \frac{\hbar \omega}{e^{\frac{\hbar \omega}{k_b T}}-1} D_p(\omega) dw\). And the expression in the denominator is just \(\displaystyle e^x-1\) after making the substitution. So we have (before taking the derivative)

\(\displaystyle \displaystyle U = \sum_p \int \frac{\hbar \omega}{e^{\frac{\hbar \omega}{k_b T}}-1} D_p(\omega) dw=\sum_p \int \frac{x k_b T}{e^x-1} D_p(\omega) dw\)

The change of variable substitution for \(\displaystyle x\) is stated to be done in the book, but only the final result is shown after taking the derivative \(\displaystyle \frac{dU}{dT}\). None of the intermediate steps are shown.

 

[Diogenes]

I was finally able to achieve the same result as the book by not using the chain rule and just taking the derivative \(\displaystyle \frac{dU}{dT}\) first, then substituting for \(\displaystyle x\) at the end. Which follows. However I am still at a loss why my 1st attempt doesn't work?


Here is what I did earlier this morning to get the result

\(\displaystyle
\begin{align*}
\frac{dU}{dT} & = \frac{d}{dT} \Biggr(\sum_p \int \frac{\hbar \omega}{e^\frac{\hbar \omega}{k_b T}-1} D_p(\omega) dw \Biggr)\\
&= \sum_p \int \frac{d}{dT} \Biggr( \frac{\hbar \omega}{e^\frac{\hbar \omega}{k_b T}-1} D_p(\omega) dw \Biggr)\\
&= \sum_p \int \hbar \omega D_p(\omega) dw \frac{d}{dT} \Biggr( \frac{1}{e^\frac{\hbar \omega}{k_b T}-1} \Biggr)\\
&= \sum_p \int \hbar \omega D_p(\omega)dw \Biggr( \frac{(0)*(e^{\frac{\hbar \omega}{k_b T}}-1) - (1)*(-\frac{\hbar \omega}{k_b T^2}e^{\frac{\hbar \omega}{k_b T}} )}{(e^{\frac{\hbar \omega}{k_b T}}-1)^2} \Biggr)\\
&= \sum_p \int \hbar \omega D_p(\omega)dw \Biggr( \frac{\frac{\hbar \omega}{k_b T^2}e^{\frac{\hbar \omega}{k_b T}} } {(e^{\frac{\hbar \omega}{k_b T}}-1)^2} \Biggr)\\
&= \sum_p \int D_p(\omega)dw \Biggr( \frac{\frac{\hbar^2 \omega^2}{k_b T^2}e^{\frac{\hbar \omega}{k_b T}} } {(e^{\frac{\hbar \omega}{k_b T}}-1)^2} \Biggr)\\
&= \sum_p \int D_p(\omega)dw \Biggr( \frac{ \frac{k_b}{k_b} \frac{\hbar^2 \omega^2}{k_b T^2}e^{\frac{\hbar \omega}{k_b T}} } {(e^{\frac{\hbar \omega}{k_b T}}-1)^2} \Biggr)\\
&= \sum_p \int D_p(\omega)dw \Biggr( \frac{ k_b \frac{\hbar^2 \omega^2}{k_b^2 T^2}e^{\frac{\hbar \omega}{k_b T}} } {(e^{\frac{\hbar \omega}{k_b T}}-1)^2} \Biggr)\\
&= \sum_p \int D_p(\omega)dw \Biggr( \frac{ k_b (\frac{\hbar \omega}{k_b T})^2 e^{\frac{\hbar \omega}{k_b T}} } {(e^{\frac{\hbar \omega}{k_b T}}-1)^2} \Biggr)\\
\end{align*}
\)


Now substituting \(\displaystyle x\) in everywhere there is an \(\displaystyle \frac{\hbar \omega}{k_b T}\)


\(\displaystyle
\begin{align*}
\frac{dU}{dT} &= \sum_p \int D_p(\omega)dw \Biggr( \frac{ k_b (\frac{\hbar \omega}{k_b T})^2 e^{\frac{\hbar \omega}{k_b T}} } {(e^{\frac{\hbar \omega}{k_b T}}-1)^2} \Biggr)\\
\frac{dU}{dT} &= \sum_p \int D_p(\omega)dw \Biggr( \frac{ k_b x^2 e^x } {(e^{\frac{\hbar \omega}{k_b T}}-1)^2} \Biggr)\\
\end{align*}
\)


This is really a much easier way and I should have thought of it first. Just do the derivative and make the \(\displaystyle x\) substitution after taking the derivative instead of making the \(\displaystyle x\) substitution before taking the derivative and using the chain rule.


However now this has really stoked my curiosity as to why my 1st attempt doesn't work at all, or, I have made some stupid mistake somewhere.


Anyone have any idea why?