5 formatting
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The following input text

  Clear[ρ, ν, λ, T, h, c, kB]
  λ[ν_] := c/ν
  ρ[ν_, T_] := (8 πh/c^3) (ν^3/(Exp[h ν/(kB T)] - 1)) \[DifferentialD]ν

  ρ[ν, T]
  ρ[λ, T]

, which visually appears as $$ \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} $$

, is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$$$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

  (8 \[Lambda]^3λ^3 \[Pi]hπh \[DifferentialD]\[Lambda]\[DifferentialD]λ)/(c^3 (-1 + E^((\[Lambda]λ)/(kB T))))

, which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals  ?

The following input text

  Clear[ρ, ν, λ, T, h, c, kB]
  λ[ν_] := c/ν
  ρ[ν_, T_] := (8 πh/c^3) (ν^3/(Exp[h ν/(kB T)] - 1)) \[DifferentialD]ν

  ρ[ν, T]
  ρ[λ, T]

, which visually appears as $$ \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} $$

, is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))

, which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals  ?

The following input text

Clear[ρ, ν, λ, T, h, c, kB]
λ[ν_] := c/ν
ρ[ν_, T_] := (8 πh/c^3) (ν^3/(Exp[h ν/(kB T)] - 1)) \[DifferentialD]ν

ρ[ν, T]
ρ[λ, T]

is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

(8 λ^3 πh \[DifferentialD]λ)/(c^3 (-1 + E^((λ)/(kB T))))

which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals?

4 Format code
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The following input text

  Clear[\[Rho]Clear[ρ, \[Nu]ν, \[Lambda]λ, T, h, c, kB]
  \[Lambda][\[Nu]_]λ[ν_] := c/\[Nu]ν
  \[Rho][\[Nu]_ρ[ν_, T_] := (8 \[Pi]hπh/
       c^3) (\[Nu]^3ν^3/(Exp[h \[Nu]ν/(kB T)] - 1)) \[DifferentialD]\[Nu]\[DifferentialD]ν

  \[Rho][\[Nu]ρ[ν, T]
  \[Rho][\[Lambda]ρ[λ, T]

, which visually appears as $$ \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} $$

, is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))

, which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ?

The following input text

  Clear[\[Rho], \[Nu], \[Lambda], T, h, c, kB]
  \[Lambda][\[Nu]_] := c/\[Nu]
  \[Rho][\[Nu]_, T_] := (8 \[Pi]h/
       c^3) (\[Nu]^3/(Exp[h \[Nu]/(kB T)] - 1)) \[DifferentialD]\[Nu]

  \[Rho][\[Nu], T]
  \[Rho][\[Lambda], T]

, which visually appears as $$ \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} $$

, is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))

, which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ?

The following input text

  Clear[ρ, ν, λ, T, h, c, kB]
  λ[ν_] := c/ν
  ρ[ν_, T_] := (8 πh/c^3) (ν^3/(Exp[h ν/(kB T)] - 1)) \[DifferentialD]ν

  ρ[ν, T]
  ρ[λ, T]

, which visually appears as $$ \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} $$

, is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))

, which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ?

3 Space tweaks.
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The following input text

  Clear[\[Rho], \[Nu], \[Lambda], T, h, c, kB]
  \[Lambda][\[Nu]_] := c/\[Nu]
  \[Rho][\[Nu]_, 
    T_] := (8 \[Pi]h/
       c^3) (\[Nu]^3/(Exp[h \[Nu]/(kB T)] - 1)) \[DifferentialD]\[Nu]

  \[Rho][\[Nu], T]
  \[Rho][\[Lambda], T]

, which visually appears as $$ \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} $$

, is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))

, which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ?

The following input text

  Clear[\[Rho], \[Nu], \[Lambda], T, h, c, kB]
  \[Lambda][\[Nu]_] := c/\[Nu]
  \[Rho][\[Nu]_, 
    T_] := (8 \[Pi]h/
       c^3) (\[Nu]^3/(Exp[h \[Nu]/(kB T)] - 1)) \[DifferentialD]\[Nu]

  \[Rho][\[Nu], T]
  \[Rho][\[Lambda], T]

, which visually appears as $$ \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} $$

, is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))

, which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ?

The following input text

  Clear[\[Rho], \[Nu], \[Lambda], T, h, c, kB]
  \[Lambda][\[Nu]_] := c/\[Nu]
  \[Rho][\[Nu]_, T_] := (8 \[Pi]h/
       c^3) (\[Nu]^3/(Exp[h \[Nu]/(kB T)] - 1)) \[DifferentialD]\[Nu]

  \[Rho][\[Nu], T]
  \[Rho][\[Lambda], T]

, which visually appears as $$ \begin{align} &\text{Clear}[\rho ,\nu ,\lambda ,T,h,c,k_\text{B}] \\ &\lambda (\nu \_):=\frac{c}{\nu } \\ &\rho (\nu \_,T\_):=\frac{8 \pi h}{c^3}\;\frac{\nu ^3 \; d\nu }{ \exp \left(\frac{h \nu }{k_\text{B} T}\right)-1} \\ &\rho [\nu ,T] \\ &\rho [\lambda ,T] \end{align} $$

, is employed to get the expression for $\rho [\lambda ,T]$, which is expected to be (equivalent to) $$ \frac{8 \pi hc}{\lambda^5}\;\frac{d\lambda }{ \exp \left(\frac{h c}{\lambda k_\text{B} T}\right)-1} $$

but turns out to be

  (8 \[Lambda]^3 \[Pi]h \[DifferentialD]\[Lambda])/(c^3 (-1 + E^((\[Lambda])/(kB T))))

, which visually appears \[ \frac{8 \lambda^3 \pi h\; d\lambda}{c^3 [\exp \left(\frac{h \lambda} {k_\text{B} T}\right)-1]} \]

As can be told by comparison, $\lambda$ is literally substituted for $\nu$. Could you help to suggest how to correctly inter-convert between two related infinitesimals ?

2 deleted 15 characters in body
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