# Integrating monochromatic luminosity to get total luminosity

## Homework Statement

Integrate the monochromatic luminosity (of a blackbody model star) over all wavelengths to obtain an expression for the total luminosity. This is 3.14(a) from An Introduction to Modern Astrophysics 2e by Carroll and Ostlie.

## Homework Equations

$L_\lambda d\lambda = \frac{8 \pi^2 R^2 h c^2 / \lambda^5}{e^{h c / \lambda k T} - 1} d\lambda$

$\int_0^\infty \frac{u^3}{e^u - 1} \, du = \frac{\pi^4}{15}$

## The Attempt at a Solution

So we need to evaluate

$L = \int_0^\infty L_\lambda \, d\lambda = \int_0^\infty \frac{8 \pi^2 R^2 h c^2 / \lambda^5}{e^{h c / \lambda k T} - 1} \, d\lambda$

$L = 8 \pi^2 R^2 h c^2 \int_0^\infty \frac{1}{\lambda^5 (e^{h c / \lambda k T} - 1)} \, d\lambda$

Using u-substitution with

$u = \frac{hc}{\lambda k T}$
$du = \frac{-hc}{\lambda^2 k T} d\lambda$

results in

$L = 8 \pi^2 R^2 h c^2 \int_0^\infty \frac{1}{e^{h c / \lambda k T} - 1} \frac{1}{\lambda^5} \, d\lambda = 8 \pi^2 R^2 h c^2 \int_0^\infty \frac{1}{e^u - 1} \frac{u^3 k^3 T^3}{h^3 c^3} \frac{-k T}{h c} \, du$

so that makes

$L = \frac{-8 \pi^2 R^2 k^4 T^4}{c^2 h^3} \int_0^\infty \frac{u^3}{e^u - 1} \, du$

and using the definition of the definite integral from the givens we get

$L = \frac{-8 \pi^6 R^2 k^4 T^4}{15 c^2 h^3}$

Everything here makes sense and is in agreement with the solution I've seen as well as the followup problems except for the negative factor. It also doesn't make much sense to have a negative luminosity in this context. I expect I've botched something small in the u-substitution but for the life of me I can't spot it.

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LCKurtz
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## Homework Statement

Integrate the monochromatic luminosity (of a blackbody model star) over all wavelengths to obtain an expression for the total luminosity. This is 3.14(a) from An Introduction to Modern Astrophysics 2e by Carroll and Ostlie.

## Homework Equations

$L_\lambda d\lambda = \frac{8 \pi^2 R^2 h c^2 / \lambda^5}{e^{h c / \lambda k T} - 1} d\lambda$

$\int_0^\infty \frac{u^3}{e^u - 1} \, du = \frac{\pi^4}{15}$

## The Attempt at a Solution

So we need to evaluate

$L = \int_0^\infty L_\lambda \, d\lambda = \int_0^\infty \frac{8 \pi^2 R^2 h c^2 / \lambda^5}{e^{h c / \lambda k T} - 1} \, d\lambda$

$L = 8 \pi^2 R^2 h c^2 \int_0^\infty \frac{1}{\lambda^5 (e^{h c / \lambda k T} - 1)} \, d\lambda$

Using u-substitution with

$u = \frac{hc}{\lambda k T}$
$du = \frac{-hc}{\lambda^2 k T} d\lambda$

results in

$L = 8 \pi^2 R^2 h c^2 \int_0^\infty \frac{1}{e^{h c / \lambda k T} - 1} \frac{1}{\lambda^5} \, d\lambda = 8 \pi^2 R^2 h c^2 \int_0^\infty \frac{1}{e^u - 1} \frac{u^3 k^3 T^3}{h^3 c^3} \frac{-k T}{h c} \, du$

Ignoring all the constants, you have an integral with limits $\int_0^\infty$ and a $\frac 1 \lambda$ and some other stuff in it. When you do a substitution like $u = \frac 1 \lambda$ and $du = -\frac1 {\lambda^2}~d\lambda$ and change the limits from $\lambda$ limits to $u$ limits, note that $\lambda = 0$ gives $u = \infty$ and $\lambda = \infty$ gives $u=0$. So you have the $u$ limits reversed and that will fix your minus sign.

• btouellette
Thank you! That's it exactly, forgot that I have to substitute out the limits as well.