Question from Advanced Engineering Mathematics - Greenberg. Page 268, section 5.4 question 6.
$C(T)$ = $\int_0^{\infty} e^{-0.0744v^2/T^2}p(v)dv$
is an approximate relation between frequency spectrum p(v) and the specific heat C(T) of a crystal, where T is the temperature.
Solve for $p(v)$ if $C(T) = T$.
Hint (given in book): By a suitable change of variables, the integral can be made to be a Laplace transform.
Spent hours on this one, got nowhere.
Looking for an appropriate change of variables to put $\int_0^{\infty} e^{-0.0744v^2/T^2}p(v)dv$ into the form:
$P(s) = \int_0^{\infty} e^{-st}p(t)dt$ - The Laplace Transform of $p(t)$. Then we can use the inverse table of the transformation to solve for $p(t)$.
Taking $s = -0.0744v/T^2$ doesn't seem to be very useful, because then the transformed function is of the form $P(-0.0744v/T^2) = T$, which is still a function of the variable $v$.
Any help is very much appreciated. Thanks.
You want $−0.0744 v^22/T^2$ to be $-s t$. Let us get rid of the square in $v^2$. We can take $u=v^2$, so $du=2v\,dv$. We substitute, and now we have $$ \int_0^\infty\,e^{−0.0744 u/T^2}\,\left(\frac1{2\sqrt u}\,p(\sqrt u)\right)\,du. $$ This is the Laplace Transform of the function in brackets at the point $s=0.0744/T^2$. So your equation is now $$ T=F(0.0744/T^2). $$ If we let $R=0.0744/T^2$, then $T=\sqrt{0.0744/R}$, and now the equation looks like $$ \frac{\sqrt{0.0744}}{\sqrt{R}}=F(R) $$ Looking at the inverse Laplace Transform of $\sqrt a/\sqrt s$, we get that $$ \frac1{2\sqrt u}\,p(\sqrt u)=\frac{\sqrt{0.0744}}{\sqrt \pi\,\sqrt u}. $$ As $\sqrt u=v$, we get $$ p(v)=\frac{2\times\sqrt{0.0744}}{\sqrt \pi}. $$ It not that exciting that $p$ turns out to be constant...