Show that $f$ is integrable with respect to the Lebesgue measure
$begingroup$
Let $alpha gt 0 forall alpha inmathbb{R}$, $f_{alpha}:]0,infty[tomathbb{R}$, $xmapsto e^{-alpha x}left(frac{sin(x)}{x}right)^3$. The goal ist to show that $f_{alpha}$ is integrable with respect to the Lebesgue-measure, which coincides with the Borel measure in this case. I already showed that $f_{alpha}$ is measurable and know that the condition for $f_{alpha}$ being integrable is that $$int_Xlvert{f(x)}rvertmathrm{d}lambdaltinfty$$ with $X=]0,infty[$. Since I am fairly new to this topic, I don't know where to start. I struggle to write out the integral since it can attain negative values and lack a general strategy for exercises like this. Any help is greatly appreciated!
real-analysis measure-theory
edited Jan 11 at 12:24
Thomas Shelby
3,7342625
asked Jan 11 at 12:13
Michael MaierMichael Maier
859
$endgroup$
add a comment |
$begingroup$
Let $alpha gt 0 forall alpha inmathbb{R}$, $f_{alpha}:]0,infty[tomathbb{R}$, $xmapsto e^{-alpha x}left(frac{sin(x)}{x}right)^3$. The goal ist to show that $f_{alpha}$ is integrable with respect to the Lebesgue-measure, which coincides with the Borel measure in this case. I already showed that $f_{alpha}$ is measurable and know that the condition for $f_{alpha}$ being integrable is that $$int_Xlvert{f(x)}rvertmathrm{d}lambdaltinfty$$ with $X=]0,infty[$. Since I am fairly new to this topic, I don't know where to start. I struggle to write out the integral since it can attain negative values and lack a general strategy for exercises like this. Any help is greatly appreciated!
real-analysis measure-theory
edited Jan 11 at 12:24
Thomas Shelby
3,7342625
asked Jan 11 at 12:13
Michael MaierMichael Maier
859
$endgroup$
$begingroup$
I guess it would be helpful to split the integral in two parts and then write it down explicitly and then to hope that the sum of my $lambda$-measures somehow converges?
$endgroup$
– Michael Maier
Jan 11 at 12:14
add a comment |
$begingroup$
Let $alpha gt 0 forall alpha inmathbb{R}$, $f_{alpha}:]0,infty[tomathbb{R}$, $xmapsto e^{-alpha x}left(frac{sin(x)}{x}right)^3$. The goal ist to show that $f_{alpha}$ is integrable with respect to the Lebesgue-measure, which coincides with the Borel measure in this case. I already showed that $f_{alpha}$ is measurable and know that the condition for $f_{alpha}$ being integrable is that $$int_Xlvert{f(x)}rvertmathrm{d}lambdaltinfty$$ with $X=]0,infty[$. Since I am fairly new to this topic, I don't know where to start. I struggle to write out the integral since it can attain negative values and lack a general strategy for exercises like this. Any help is greatly appreciated!
real-analysis measure-theory
edited Jan 11 at 12:24
Thomas Shelby
3,7342625
asked Jan 11 at 12:13
Michael MaierMichael Maier
859
$endgroup$
Let $alpha gt 0 forall alpha inmathbb{R}$, $f_{alpha}:]0,infty[tomathbb{R}$, $xmapsto e^{-alpha x}left(frac{sin(x)}{x}right)^3$. The goal ist to show that $f_{alpha}$ is integrable with respect to the Lebesgue-measure, which coincides with the Borel measure in this case. I already showed that $f_{alpha}$ is measurable and know that the condition for $f_{alpha}$ being integrable is that $$int_Xlvert{f(x)}rvertmathrm{d}lambdaltinfty$$ with $X=]0,infty[$. Since I am fairly new to this topic, I don't know where to start. I struggle to write out the integral since it can attain negative values and lack a general strategy for exercises like this. Any help is greatly appreciated!
real-analysis measure-theory
real-analysis measure-theory
edited Jan 11 at 12:24
Thomas Shelby
3,7342625
asked Jan 11 at 12:13
Michael MaierMichael Maier
859
edited Jan 11 at 12:24
Thomas Shelby
3,7342625
asked Jan 11 at 12:13
Michael MaierMichael Maier
859
edited Jan 11 at 12:24
Thomas Shelby
3,7342625
edited Jan 11 at 12:24
Thomas Shelby
3,7342625
edited Jan 11 at 12:24
Thomas Shelby
3,7342625
3,7342625
asked Jan 11 at 12:13
Michael MaierMichael Maier
859
asked Jan 11 at 12:13
Michael MaierMichael Maier
859
asked Jan 11 at 12:13
Michael MaierMichael Maier
859
859
$begingroup$
I guess it would be helpful to split the integral in two parts and then write it down explicitly and then to hope that the sum of my $lambda$-measures somehow converges?
$endgroup$
– Michael Maier
Jan 11 at 12:14
add a comment |
$begingroup$
I guess it would be helpful to split the integral in two parts and then write it down explicitly and then to hope that the sum of my $lambda$-measures somehow converges?
$endgroup$
– Michael Maier
Jan 11 at 12:14
$begingroup$
I guess it would be helpful to split the integral in two parts and then write it down explicitly and then to hope that the sum of my $lambda$-measures somehow converges?
$endgroup$
– Michael Maier
Jan 11 at 12:14
$begingroup$
I guess it would be helpful to split the integral in two parts and then write it down explicitly and then to hope that the sum of my $lambda$-measures somehow converges?
$endgroup$
– Michael Maier
Jan 11 at 12:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$frac {sin, x} x$ is abounded function on $(0,infty)$. If $|frac {sin, x} x| leq M$ then $int_0^{infty} |f_{alpha} (x)| , dx leq M^{3}int_0^{infty} e^{-alpha x}, dx =frac {M^{3}} {alpha}$. [ Boundedness of $frac {sin, x} x$ is a consequence of the following facts: this function is continuous, it approaches $1$ as $ xto 0$ and approaches $0$ as $x to infty$].
answered Jan 11 at 12:33
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
Thanks. Problem is, we didn't cover $int_0^infty e^{-alpha x}dx=frac{1}{alpha}$. I just need to show that $int_0^infty e^{-alpha x}dlambda$ is integrable then, any advice on that? Obviously it is decreasing, but I don't see where that might help.
$endgroup$
– Michael Maier
Jan 11 at 12:37
1
$begingroup$
Surely if you're doing Lebesgue measure, you know how to evaluate a definite integral. Are you required to use only things covered in the course?
$endgroup$
– Riccardo Orlando
Jan 11 at 12:38
1
$begingroup$
@MichaelMaier Riemann integrable functions are Lebesgue integrable and their Lebesgue integral is same as Riemann integral. I am sure you have covered intergral of $e^{-alpha x}$ before starting with measure theory.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:45
$begingroup$
Thanks again. As a physics student, I indeed know what the integral of $e^{-alpha x}$ is. Problem is, our math lectures are unrelated to that and we do just mention Riemann integrals later on. So I do know how to calculate the integral, but am not allowed to use anything related to Riemannian Integration in my proof.
$endgroup$
– Michael Maier
Jan 11 at 12:48
1
$begingroup$
@MichaelMaier If $a_n=frac 2 {alpha} log, n$ then $a_n$ increases to $infty$. Split the integral of $e^{-alpha x}$ into integrals over the intervals $(a_n,a_{n+1})$. Can use this to show that the integral is finite? (On $(a_n,a_{n+1})$ note that $e^{-alpha x} leq e^{-alpha a_n} =frac 1 {n^{2}}$).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:49
|
show 2 more comments
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$begingroup$
$frac {sin, x} x$ is abounded function on $(0,infty)$. If $|frac {sin, x} x| leq M$ then $int_0^{infty} |f_{alpha} (x)| , dx leq M^{3}int_0^{infty} e^{-alpha x}, dx =frac {M^{3}} {alpha}$. [ Boundedness of $frac {sin, x} x$ is a consequence of the following facts: this function is continuous, it approaches $1$ as $ xto 0$ and approaches $0$ as $x to infty$].
answered Jan 11 at 12:33
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
Thanks. Problem is, we didn't cover $int_0^infty e^{-alpha x}dx=frac{1}{alpha}$. I just need to show that $int_0^infty e^{-alpha x}dlambda$ is integrable then, any advice on that? Obviously it is decreasing, but I don't see where that might help.
$endgroup$
– Michael Maier
Jan 11 at 12:37
1
$begingroup$
Surely if you're doing Lebesgue measure, you know how to evaluate a definite integral. Are you required to use only things covered in the course?
$endgroup$
– Riccardo Orlando
Jan 11 at 12:38
1
$begingroup$
@MichaelMaier Riemann integrable functions are Lebesgue integrable and their Lebesgue integral is same as Riemann integral. I am sure you have covered intergral of $e^{-alpha x}$ before starting with measure theory.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:45
$begingroup$
Thanks again. As a physics student, I indeed know what the integral of $e^{-alpha x}$ is. Problem is, our math lectures are unrelated to that and we do just mention Riemann integrals later on. So I do know how to calculate the integral, but am not allowed to use anything related to Riemannian Integration in my proof.
$endgroup$
– Michael Maier
Jan 11 at 12:48
1
$begingroup$
@MichaelMaier If $a_n=frac 2 {alpha} log, n$ then $a_n$ increases to $infty$. Split the integral of $e^{-alpha x}$ into integrals over the intervals $(a_n,a_{n+1})$. Can use this to show that the integral is finite? (On $(a_n,a_{n+1})$ note that $e^{-alpha x} leq e^{-alpha a_n} =frac 1 {n^{2}}$).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:49
|
show 2 more comments
$begingroup$
$frac {sin, x} x$ is abounded function on $(0,infty)$. If $|frac {sin, x} x| leq M$ then $int_0^{infty} |f_{alpha} (x)| , dx leq M^{3}int_0^{infty} e^{-alpha x}, dx =frac {M^{3}} {alpha}$. [ Boundedness of $frac {sin, x} x$ is a consequence of the following facts: this function is continuous, it approaches $1$ as $ xto 0$ and approaches $0$ as $x to infty$].
answered Jan 11 at 12:33
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
Thanks. Problem is, we didn't cover $int_0^infty e^{-alpha x}dx=frac{1}{alpha}$. I just need to show that $int_0^infty e^{-alpha x}dlambda$ is integrable then, any advice on that? Obviously it is decreasing, but I don't see where that might help.
$endgroup$
– Michael Maier
Jan 11 at 12:37
1
$begingroup$
Surely if you're doing Lebesgue measure, you know how to evaluate a definite integral. Are you required to use only things covered in the course?
$endgroup$
– Riccardo Orlando
Jan 11 at 12:38
1
$begingroup$
@MichaelMaier Riemann integrable functions are Lebesgue integrable and their Lebesgue integral is same as Riemann integral. I am sure you have covered intergral of $e^{-alpha x}$ before starting with measure theory.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:45
$begingroup$
Thanks again. As a physics student, I indeed know what the integral of $e^{-alpha x}$ is. Problem is, our math lectures are unrelated to that and we do just mention Riemann integrals later on. So I do know how to calculate the integral, but am not allowed to use anything related to Riemannian Integration in my proof.
$endgroup$
– Michael Maier
Jan 11 at 12:48
1
$begingroup$
@MichaelMaier If $a_n=frac 2 {alpha} log, n$ then $a_n$ increases to $infty$. Split the integral of $e^{-alpha x}$ into integrals over the intervals $(a_n,a_{n+1})$. Can use this to show that the integral is finite? (On $(a_n,a_{n+1})$ note that $e^{-alpha x} leq e^{-alpha a_n} =frac 1 {n^{2}}$).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:49
|
show 2 more comments
$begingroup$
$frac {sin, x} x$ is abounded function on $(0,infty)$. If $|frac {sin, x} x| leq M$ then $int_0^{infty} |f_{alpha} (x)| , dx leq M^{3}int_0^{infty} e^{-alpha x}, dx =frac {M^{3}} {alpha}$. [ Boundedness of $frac {sin, x} x$ is a consequence of the following facts: this function is continuous, it approaches $1$ as $ xto 0$ and approaches $0$ as $x to infty$].
answered Jan 11 at 12:33
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$frac {sin, x} x$ is abounded function on $(0,infty)$. If $|frac {sin, x} x| leq M$ then $int_0^{infty} |f_{alpha} (x)| , dx leq M^{3}int_0^{infty} e^{-alpha x}, dx =frac {M^{3}} {alpha}$. [ Boundedness of $frac {sin, x} x$ is a consequence of the following facts: this function is continuous, it approaches $1$ as $ xto 0$ and approaches $0$ as $x to infty$].
answered Jan 11 at 12:33
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered Jan 11 at 12:33
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered Jan 11 at 12:33
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered Jan 11 at 12:33
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
65.6k42766
$begingroup$
Thanks. Problem is, we didn't cover $int_0^infty e^{-alpha x}dx=frac{1}{alpha}$. I just need to show that $int_0^infty e^{-alpha x}dlambda$ is integrable then, any advice on that? Obviously it is decreasing, but I don't see where that might help.
$endgroup$
– Michael Maier
Jan 11 at 12:37
1
$begingroup$
Surely if you're doing Lebesgue measure, you know how to evaluate a definite integral. Are you required to use only things covered in the course?
$endgroup$
– Riccardo Orlando
Jan 11 at 12:38
1
$begingroup$
@MichaelMaier Riemann integrable functions are Lebesgue integrable and their Lebesgue integral is same as Riemann integral. I am sure you have covered intergral of $e^{-alpha x}$ before starting with measure theory.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:45
$begingroup$
Thanks again. As a physics student, I indeed know what the integral of $e^{-alpha x}$ is. Problem is, our math lectures are unrelated to that and we do just mention Riemann integrals later on. So I do know how to calculate the integral, but am not allowed to use anything related to Riemannian Integration in my proof.
$endgroup$
– Michael Maier
Jan 11 at 12:48
1
$begingroup$
@MichaelMaier If $a_n=frac 2 {alpha} log, n$ then $a_n$ increases to $infty$. Split the integral of $e^{-alpha x}$ into integrals over the intervals $(a_n,a_{n+1})$. Can use this to show that the integral is finite? (On $(a_n,a_{n+1})$ note that $e^{-alpha x} leq e^{-alpha a_n} =frac 1 {n^{2}}$).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:49
|
show 2 more comments
$begingroup$
Thanks. Problem is, we didn't cover $int_0^infty e^{-alpha x}dx=frac{1}{alpha}$. I just need to show that $int_0^infty e^{-alpha x}dlambda$ is integrable then, any advice on that? Obviously it is decreasing, but I don't see where that might help.
$endgroup$
– Michael Maier
Jan 11 at 12:37
1
$begingroup$
Surely if you're doing Lebesgue measure, you know how to evaluate a definite integral. Are you required to use only things covered in the course?
$endgroup$
– Riccardo Orlando
Jan 11 at 12:38
1
$begingroup$
@MichaelMaier Riemann integrable functions are Lebesgue integrable and their Lebesgue integral is same as Riemann integral. I am sure you have covered intergral of $e^{-alpha x}$ before starting with measure theory.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:45
$begingroup$
Thanks again. As a physics student, I indeed know what the integral of $e^{-alpha x}$ is. Problem is, our math lectures are unrelated to that and we do just mention Riemann integrals later on. So I do know how to calculate the integral, but am not allowed to use anything related to Riemannian Integration in my proof.
$endgroup$
– Michael Maier
Jan 11 at 12:48
1
$begingroup$
@MichaelMaier If $a_n=frac 2 {alpha} log, n$ then $a_n$ increases to $infty$. Split the integral of $e^{-alpha x}$ into integrals over the intervals $(a_n,a_{n+1})$. Can use this to show that the integral is finite? (On $(a_n,a_{n+1})$ note that $e^{-alpha x} leq e^{-alpha a_n} =frac 1 {n^{2}}$).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:49
$begingroup$
Thanks. Problem is, we didn't cover $int_0^infty e^{-alpha x}dx=frac{1}{alpha}$. I just need to show that $int_0^infty e^{-alpha x}dlambda$ is integrable then, any advice on that? Obviously it is decreasing, but I don't see where that might help.
$endgroup$
– Michael Maier
Jan 11 at 12:37
$begingroup$
Thanks. Problem is, we didn't cover $int_0^infty e^{-alpha x}dx=frac{1}{alpha}$. I just need to show that $int_0^infty e^{-alpha x}dlambda$ is integrable then, any advice on that? Obviously it is decreasing, but I don't see where that might help.
$endgroup$
– Michael Maier
Jan 11 at 12:37
1
1
$begingroup$
Surely if you're doing Lebesgue measure, you know how to evaluate a definite integral. Are you required to use only things covered in the course?
$endgroup$
– Riccardo Orlando
Jan 11 at 12:38
$begingroup$
Surely if you're doing Lebesgue measure, you know how to evaluate a definite integral. Are you required to use only things covered in the course?
$endgroup$
– Riccardo Orlando
Jan 11 at 12:38
1
1
$begingroup$
@MichaelMaier Riemann integrable functions are Lebesgue integrable and their Lebesgue integral is same as Riemann integral. I am sure you have covered intergral of $e^{-alpha x}$ before starting with measure theory.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:45
$begingroup$
@MichaelMaier Riemann integrable functions are Lebesgue integrable and their Lebesgue integral is same as Riemann integral. I am sure you have covered intergral of $e^{-alpha x}$ before starting with measure theory.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:45
$begingroup$
Thanks again. As a physics student, I indeed know what the integral of $e^{-alpha x}$ is. Problem is, our math lectures are unrelated to that and we do just mention Riemann integrals later on. So I do know how to calculate the integral, but am not allowed to use anything related to Riemannian Integration in my proof.
$endgroup$
– Michael Maier
Jan 11 at 12:48
$begingroup$
Thanks again. As a physics student, I indeed know what the integral of $e^{-alpha x}$ is. Problem is, our math lectures are unrelated to that and we do just mention Riemann integrals later on. So I do know how to calculate the integral, but am not allowed to use anything related to Riemannian Integration in my proof.
$endgroup$
– Michael Maier
Jan 11 at 12:48
1
1
$begingroup$
@MichaelMaier If $a_n=frac 2 {alpha} log, n$ then $a_n$ increases to $infty$. Split the integral of $e^{-alpha x}$ into integrals over the intervals $(a_n,a_{n+1})$. Can use this to show that the integral is finite? (On $(a_n,a_{n+1})$ note that $e^{-alpha x} leq e^{-alpha a_n} =frac 1 {n^{2}}$).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:49
$begingroup$
@MichaelMaier If $a_n=frac 2 {alpha} log, n$ then $a_n$ increases to $infty$. Split the integral of $e^{-alpha x}$ into integrals over the intervals $(a_n,a_{n+1})$. Can use this to show that the integral is finite? (On $(a_n,a_{n+1})$ note that $e^{-alpha x} leq e^{-alpha a_n} =frac 1 {n^{2}}$).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 12:49
|
show 2 more comments
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$begingroup$
I guess it would be helpful to split the integral in two parts and then write it down explicitly and then to hope that the sum of my $lambda$-measures somehow converges?
$endgroup$
– Michael Maier
Jan 11 at 12:14