Convergence of $int_0^infty sin(x^m)/x^n dx$
2
$begingroup$
$$int_0^infty frac{sin (x^m)}{x^n}dx $$
Putting $x^m = t$
$$
begin{align}
frac{1}{m}int_0^infty frac{sin t}{t^{(frac{m+n-1}{m})}}dt
end{align}
$$
By applying Dirichlet Test I've been able to prove that the integrand is convergent when $1-m < n$.
But the answer given is $1-m < n < m+1$. Can you give me any hint on how to get the second condition?
real-analysis improper-integrals
$endgroup$
add a comment |
2
$begingroup$
$$int_0^infty frac{sin (x^m)}{x^n}dx $$
Putting $x^m = t$
$$
begin{align}
frac{1}{m}int_0^infty frac{sin t}{t^{(frac{m+n-1}{m})}}dt
end{align}
$$
By applying Dirichlet Test I've been able to prove that the integrand is convergent when $1-m < n$.
But the answer given is $1-m < n < m+1$. Can you give me any hint on how to get the second condition?
real-analysis improper-integrals
$endgroup$
1
$begingroup$
Were the limits in the actual problem from $0$ to $infty$ (not $1$ to $infty$)?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 4:15
1
$begingroup$
If $m>0$, then we must have $n>1-m$ for convergence.
$endgroup$
– Mark Viola
Dec 14 '18 at 4:34
$begingroup$
@Mark Viola That I've been able to show. How to prove $n<m+1$ ?
$endgroup$
– user626133
Dec 14 '18 at 17:04
$begingroup$
$n<m+1$ cannot possibly be true. If it converges for $n$ given $m$, then it converges for $n+1, n+2,...$ etc.
$endgroup$
– dezdichado
Dec 14 '18 at 18:02
$begingroup$
There was a mistake in the question the bounds of integration are from $0$ to $infty$. Sorry, it was a typo by the TA. I'm able to solve the problem and will post the solution also. Thanks for your support.
$endgroup$
– user626133
Dec 14 '18 at 19:05
add a comment |
2
2
2
$begingroup$
$$int_0^infty frac{sin (x^m)}{x^n}dx $$
Putting $x^m = t$
$$
begin{align}
frac{1}{m}int_0^infty frac{sin t}{t^{(frac{m+n-1}{m})}}dt
end{align}
$$
By applying Dirichlet Test I've been able to prove that the integrand is convergent when $1-m < n$.
But the answer given is $1-m < n < m+1$. Can you give me any hint on how to get the second condition?
real-analysis improper-integrals
$endgroup$
$$int_0^infty frac{sin (x^m)}{x^n}dx $$
Putting $x^m = t$
$$
begin{align}
frac{1}{m}int_0^infty frac{sin t}{t^{(frac{m+n-1}{m})}}dt
end{align}
$$
By applying Dirichlet Test I've been able to prove that the integrand is convergent when $1-m < n$.
But the answer given is $1-m < n < m+1$. Can you give me any hint on how to get the second condition?
real-analysis improper-integrals
real-analysis improper-integrals
edited Dec 14 '18 at 21:00
asked Dec 14 '18 at 4:06
user626133
1
$begingroup$
Were the limits in the actual problem from $0$ to $infty$ (not $1$ to $infty$)?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 4:15
1
$begingroup$
If $m>0$, then we must have $n>1-m$ for convergence.
$endgroup$
– Mark Viola
Dec 14 '18 at 4:34
$begingroup$
@Mark Viola That I've been able to show. How to prove $n<m+1$ ?
$endgroup$
– user626133
Dec 14 '18 at 17:04
$begingroup$
$n<m+1$ cannot possibly be true. If it converges for $n$ given $m$, then it converges for $n+1, n+2,...$ etc.
$endgroup$
– dezdichado
Dec 14 '18 at 18:02
$begingroup$
There was a mistake in the question the bounds of integration are from $0$ to $infty$. Sorry, it was a typo by the TA. I'm able to solve the problem and will post the solution also. Thanks for your support.
$endgroup$
– user626133
Dec 14 '18 at 19:05
add a comment |
1
$begingroup$
Were the limits in the actual problem from $0$ to $infty$ (not $1$ to $infty$)?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 4:15
1
$begingroup$
If $m>0$, then we must have $n>1-m$ for convergence.
$endgroup$
– Mark Viola
Dec 14 '18 at 4:34
$begingroup$
@Mark Viola That I've been able to show. How to prove $n<m+1$ ?
$endgroup$
– user626133
Dec 14 '18 at 17:04
$begingroup$
$n<m+1$ cannot possibly be true. If it converges for $n$ given $m$, then it converges for $n+1, n+2,...$ etc.
$endgroup$
– dezdichado
Dec 14 '18 at 18:02
$begingroup$
There was a mistake in the question the bounds of integration are from $0$ to $infty$. Sorry, it was a typo by the TA. I'm able to solve the problem and will post the solution also. Thanks for your support.
$endgroup$
– user626133
Dec 14 '18 at 19:05
1
1
$begingroup$
Were the limits in the actual problem from $0$ to $infty$ (not $1$ to $infty$)?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 4:15
$begingroup$
Were the limits in the actual problem from $0$ to $infty$ (not $1$ to $infty$)?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 4:15
1
1
$begingroup$
If $m>0$, then we must have $n>1-m$ for convergence.
$endgroup$
– Mark Viola
Dec 14 '18 at 4:34
$begingroup$
If $m>0$, then we must have $n>1-m$ for convergence.
$endgroup$
– Mark Viola
Dec 14 '18 at 4:34
$begingroup$
@Mark Viola That I've been able to show. How to prove $n<m+1$ ?
$endgroup$
– user626133
Dec 14 '18 at 17:04
$begingroup$
@Mark Viola That I've been able to show. How to prove $n<m+1$ ?
$endgroup$
– user626133
Dec 14 '18 at 17:04
$begingroup$
$n<m+1$ cannot possibly be true. If it converges for $n$ given $m$, then it converges for $n+1, n+2,...$ etc.
$endgroup$
– dezdichado
Dec 14 '18 at 18:02
$begingroup$
$n<m+1$ cannot possibly be true. If it converges for $n$ given $m$, then it converges for $n+1, n+2,...$ etc.
$endgroup$
– dezdichado
Dec 14 '18 at 18:02
$begingroup$
There was a mistake in the question the bounds of integration are from $0$ to $infty$. Sorry, it was a typo by the TA. I'm able to solve the problem and will post the solution also. Thanks for your support.
$endgroup$
– user626133
Dec 14 '18 at 19:05
$begingroup$
There was a mistake in the question the bounds of integration are from $0$ to $infty$. Sorry, it was a typo by the TA. I'm able to solve the problem and will post the solution also. Thanks for your support.
$endgroup$
– user626133
Dec 14 '18 at 19:05
add a comment |
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1
$begingroup$
Were the limits in the actual problem from $0$ to $infty$ (not $1$ to $infty$)?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 4:15
1
$begingroup$
If $m>0$, then we must have $n>1-m$ for convergence.
$endgroup$
– Mark Viola
Dec 14 '18 at 4:34
$begingroup$
@Mark Viola That I've been able to show. How to prove $n<m+1$ ?
$endgroup$
– user626133
Dec 14 '18 at 17:04
$begingroup$
$n<m+1$ cannot possibly be true. If it converges for $n$ given $m$, then it converges for $n+1, n+2,...$ etc.
$endgroup$
– dezdichado
Dec 14 '18 at 18:02
$begingroup$
There was a mistake in the question the bounds of integration are from $0$ to $infty$. Sorry, it was a typo by the TA. I'm able to solve the problem and will post the solution also. Thanks for your support.
$endgroup$
– user626133
Dec 14 '18 at 19:05