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Mathematics
Calculus
Integrals over chained functions
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[QUOTE="SchroedingersLion, post: 6173808, member: 624040"] Good evening! Going through a bunch of calculations in Ashcroft's and Mermin's Solid State Physics, I have come across either an error on their part or a missunderstanding on my part. Suppose we have a concatenated function, say the fermi function ##f(\epsilon)## that goes from R to R. We know that ##\epsilon## is a function of the 3 dimensional wave vector [B]k[/B] . If there are applied electromagnetic fields, the wave vector depends on time t', i.e. [B]k[/B](t')[B].[/B] Now suppose we want to calculate an integral $$\int_{-\infty}^t p(t')\frac {df}{dt'} \, dt' $$ with some scalar function ##p(t')##. Chain rule tells us that ##\frac {df}{dt'} = \frac {df}{d\epsilon} \frac {d\epsilon}{d\mathbf k}\frac {d\mathbf k}{dt'}##, so we get $$ \int_{-\infty}^t p(t')\frac {df}{d\epsilon} \frac {d\epsilon}{d\mathbf k}\frac {d\mathbf k}{dt'} \, dt' $$ So, assuming that ##\frac {df}{d\epsilon}## would still depend on ##\epsilon ## and ##\frac {d\epsilon}{d\mathbf k}## would still depend on ##\mathbf k##, am I right in assuming that both of these terms would also be considered as being dependent on t'? Because ##\epsilon## depends on ##\mathbf k## and the latter on t'. So, under the integral, I would express ##\epsilon## and ##\mathbf k## in terms of t', before I integrate. Yet, the book keeps acting as if the time dependence only comes from ##p(t)## and ##\frac {d\mathbf k}{dt'}##.SL [/QUOTE]
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Mathematics
Calculus
Integrals over chained functions
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