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I have a passband filter $b$ and a spectrum (SED) $f_{\lambda}(\lambda)$. Often the formula I find to get the flux in that filter $F_b$ is $$F_{b}=\dfrac{\int f_{\lambda}(\lambda) T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)d\lambda}\,,$$ where $T_b$ is the filter response in band $b$, which is similar to the filters shown in figure 2 of The PLAsTiCC team et al (2018).

However, when I try to compute a apparent magnitude, I get the formula $$m_b=-2.5\log_{10}\left(\dfrac{\int f_{\lambda}(\lambda)\,\lambda\,T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)\,\lambda\,d\lambda}\right)\,.$$

I don't understand from where does the $\lambda$ in the second formula arises, since I thought that $$m_b=-2.5\log_{10}\left(\dfrac{F_b}{F_{\mathrm{reference}}}\right)\,.$$

For example, Casagrande & VandenBerg (2014) show both expressions in equations 1 and 3. However Bessell & Murphy (2012) (eq. 2) and Hogg, Baldry, Blanton, & Eisenstein (2002) (eq. 4) only show the second approach. Additionally, Hogg (2000) shows the K-correction without the added $\lambda$'s: $$K=-2.5\log_{10}\left(\dfrac{1}{(1+z)}\dfrac{L_{\lambda/(1+z)}}{L_{\lambda}}\right)\,.$$

I don't know how to reconcile these different views. Can you help me?

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  • $\begingroup$ @usernumber, I have fixed the link. When I feed equation (1) into equation (3), the $\lambda$ factor does not appear. That is why I am confused. $\endgroup$ – Catarina Alves Feb 25 at 16:40
  • $\begingroup$ @usernumber there's an additional lambda in the denominator's integral $\endgroup$ – planetmaker Feb 25 at 17:05

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