Interchange of a sum and integral
$begingroup$
I have to compute the following integral:
$$N = -frac{2pi A}{h^2}int_{0}^{infty} ln(1-ze^{-beta pc})pmathrm{d}p$$
Let's say $z << 1$. Thus:
$$ln(1-x) = -sum_{n=1}^{infty} frac{x^n}{n}$$
$$N = frac{2pi A}{h^2}int_{0}^{infty} sum_{n=1}^{infty} mathrm{d}pfrac{z^n e^{-n beta pc}}{n}p$$
Now it is convenient to use the following change of variables:
$$x = beta pc$$
$$N = frac{2pi A}{(h beta c)^2}int_{0}^{infty} sum_{n=1}^{infty} mathrm{d}xfrac{z^n e^{-n x}}{n}x$$
It's time to factorize the series:
$$N = frac{2pi A}{(h beta c)^2}sum_{n=1}^{infty} frac{z^n}{n} int_{0}^{infty} xe^{-x} mathrm{d}x$$
But I saw in the provided solution that I am missing a factor of three on the series, as it is stated to be:
$$N = frac{2pi A}{(h beta c)^2}sum_{n=1}^{infty} frac{z^n}{n^3} int_{0}^{infty} xe^{-x} mathrm{d}x$$
From where is this to the power of 3 coming?
calculus power-series
edited Dec 1 '18 at 17:13
Daniele Tampieri
2,1191621
asked Dec 1 '18 at 16:36
JD_PMJD_PM
10410
$endgroup$
add a comment |
$begingroup$
I have to compute the following integral:
$$N = -frac{2pi A}{h^2}int_{0}^{infty} ln(1-ze^{-beta pc})pmathrm{d}p$$
Let's say $z << 1$. Thus:
$$ln(1-x) = -sum_{n=1}^{infty} frac{x^n}{n}$$
$$N = frac{2pi A}{h^2}int_{0}^{infty} sum_{n=1}^{infty} mathrm{d}pfrac{z^n e^{-n beta pc}}{n}p$$
Now it is convenient to use the following change of variables:
$$x = beta pc$$
$$N = frac{2pi A}{(h beta c)^2}int_{0}^{infty} sum_{n=1}^{infty} mathrm{d}xfrac{z^n e^{-n x}}{n}x$$
It's time to factorize the series:
$$N = frac{2pi A}{(h beta c)^2}sum_{n=1}^{infty} frac{z^n}{n} int_{0}^{infty} xe^{-x} mathrm{d}x$$
But I saw in the provided solution that I am missing a factor of three on the series, as it is stated to be:
$$N = frac{2pi A}{(h beta c)^2}sum_{n=1}^{infty} frac{z^n}{n^3} int_{0}^{infty} xe^{-x} mathrm{d}x$$
From where is this to the power of 3 coming?
calculus power-series
edited Dec 1 '18 at 17:13
Daniele Tampieri
2,1191621
asked Dec 1 '18 at 16:36
JD_PMJD_PM
10410
$endgroup$
add a comment |
$begingroup$
I have to compute the following integral:
$$N = -frac{2pi A}{h^2}int_{0}^{infty} ln(1-ze^{-beta pc})pmathrm{d}p$$
Let's say $z << 1$. Thus:
$$ln(1-x) = -sum_{n=1}^{infty} frac{x^n}{n}$$
$$N = frac{2pi A}{h^2}int_{0}^{infty} sum_{n=1}^{infty} mathrm{d}pfrac{z^n e^{-n beta pc}}{n}p$$
Now it is convenient to use the following change of variables:
$$x = beta pc$$
$$N = frac{2pi A}{(h beta c)^2}int_{0}^{infty} sum_{n=1}^{infty} mathrm{d}xfrac{z^n e^{-n x}}{n}x$$
It's time to factorize the series:
$$N = frac{2pi A}{(h beta c)^2}sum_{n=1}^{infty} frac{z^n}{n} int_{0}^{infty} xe^{-x} mathrm{d}x$$
But I saw in the provided solution that I am missing a factor of three on the series, as it is stated to be:
$$N = frac{2pi A}{(h beta c)^2}sum_{n=1}^{infty} frac{z^n}{n^3} int_{0}^{infty} xe^{-x} mathrm{d}x$$
From where is this to the power of 3 coming?
calculus power-series
edited Dec 1 '18 at 17:13
Daniele Tampieri
2,1191621
asked Dec 1 '18 at 16:36
JD_PMJD_PM
10410
$endgroup$
I have to compute the following integral:
$$N = -frac{2pi A}{h^2}int_{0}^{infty} ln(1-ze^{-beta pc})pmathrm{d}p$$
Let's say $z << 1$. Thus:
$$ln(1-x) = -sum_{n=1}^{infty} frac{x^n}{n}$$
$$N = frac{2pi A}{h^2}int_{0}^{infty} sum_{n=1}^{infty} mathrm{d}pfrac{z^n e^{-n beta pc}}{n}p$$
Now it is convenient to use the following change of variables:
$$x = beta pc$$
$$N = frac{2pi A}{(h beta c)^2}int_{0}^{infty} sum_{n=1}^{infty} mathrm{d}xfrac{z^n e^{-n x}}{n}x$$
It's time to factorize the series:
$$N = frac{2pi A}{(h beta c)^2}sum_{n=1}^{infty} frac{z^n}{n} int_{0}^{infty} xe^{-x} mathrm{d}x$$
But I saw in the provided solution that I am missing a factor of three on the series, as it is stated to be:
$$N = frac{2pi A}{(h beta c)^2}sum_{n=1}^{infty} frac{z^n}{n^3} int_{0}^{infty} xe^{-x} mathrm{d}x$$
From where is this to the power of 3 coming?
calculus power-series
calculus power-series
edited Dec 1 '18 at 17:13
Daniele Tampieri
2,1191621
asked Dec 1 '18 at 16:36
JD_PMJD_PM
10410
edited Dec 1 '18 at 17:13
Daniele Tampieri
2,1191621
asked Dec 1 '18 at 16:36
JD_PMJD_PM
10410
edited Dec 1 '18 at 17:13
Daniele Tampieri
2,1191621
edited Dec 1 '18 at 17:13
Daniele Tampieri
2,1191621
edited Dec 1 '18 at 17:13
Daniele Tampieri
2,1191621
2,1191621
asked Dec 1 '18 at 16:36
JD_PMJD_PM
10410
asked Dec 1 '18 at 16:36
JD_PMJD_PM
10410
asked Dec 1 '18 at 16:36
JD_PMJD_PM
10410
10410
add a comment |
add a comment |
1 Answer
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$begingroup$
The issue is that you went from $e^{-nx}$ to $e^{-x}$ without doing a proper change of variable. If $y=nx$ then the $x$ that you have next to the exponential becomes $y/n$ and $dx$ becomes $dy/n$. Therefore you have two extra $1/n$ factors
answered Dec 1 '18 at 17:27
AndreiAndrei
11.7k21026
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The issue is that you went from $e^{-nx}$ to $e^{-x}$ without doing a proper change of variable. If $y=nx$ then the $x$ that you have next to the exponential becomes $y/n$ and $dx$ becomes $dy/n$. Therefore you have two extra $1/n$ factors
answered Dec 1 '18 at 17:27
AndreiAndrei
11.7k21026
$endgroup$
add a comment |
$begingroup$
The issue is that you went from $e^{-nx}$ to $e^{-x}$ without doing a proper change of variable. If $y=nx$ then the $x$ that you have next to the exponential becomes $y/n$ and $dx$ becomes $dy/n$. Therefore you have two extra $1/n$ factors
answered Dec 1 '18 at 17:27
AndreiAndrei
11.7k21026
$endgroup$
add a comment |
$begingroup$
The issue is that you went from $e^{-nx}$ to $e^{-x}$ without doing a proper change of variable. If $y=nx$ then the $x$ that you have next to the exponential becomes $y/n$ and $dx$ becomes $dy/n$. Therefore you have two extra $1/n$ factors
answered Dec 1 '18 at 17:27
AndreiAndrei
11.7k21026
$endgroup$
The issue is that you went from $e^{-nx}$ to $e^{-x}$ without doing a proper change of variable. If $y=nx$ then the $x$ that you have next to the exponential becomes $y/n$ and $dx$ becomes $dy/n$. Therefore you have two extra $1/n$ factors
answered Dec 1 '18 at 17:27
AndreiAndrei
11.7k21026
edited Dec 1 '18 at 23:07
answered Dec 1 '18 at 17:27
AndreiAndrei
11.7k21026
answered Dec 1 '18 at 17:27
AndreiAndrei
11.7k21026
answered Dec 1 '18 at 17:27
AndreiAndrei
11.7k21026
11.7k21026
add a comment |
add a comment |
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