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Find $lim_{nrightarrowinfty}int_{2npi}^{2(n+1)pi}xln xcos x,dx$

Find the limit : $limlimits_{nrightarrowinfty}int_{n}^{n+7}frac{sin{x}}{x},mathrm dx$Show that $lim_{nrightarrow infty} int_0^{pi/2} 2^n sqrt{n} sin^n(x) cos^{n-2}(x) ; dx = sqrt{2pi}$The value of $lim_{nto infty}int_{-infty}^{infty}f(x)cos^{2} nx dx.$Could we solve $int_{0}^{infty}sin(x)dx$ and what does it say about $lim_{xtoinfty}cos(x)$?Integral $lim_{krightarrowinfty}int_pi^{2pi}{sum_{n=1}^k}left(sin xright)^n text{dx}$Computing $lim_{n rightarrow infty} int_{0}^{pi/3} frac{sin^{n}x}{sin^{n}x+cos^{n}x}dx$ using Dominated Convergence TheoremEvaluating $lim_{nrightarrowinfty} int_{0}^{pi} frac {sin x}{1+ cos^2 (nx)} dx$evaluating $lim_{x rightarrow 0} frac{int_{0}^{2sin x} cos(t^2) dt}{2x}$Find $lim_{xrightarrow{infty}} frac{int_{0}^{1} (1+(xy)^{n})^{frac{1}{n}}dy}{x}$, where $ngeq 2$.Find $lim_{nrightarrowinfty}sum_{k=1}^{n}frac{n}{k(2n-k+1)}$ and $lim_{nrightarrowinfty}sum_{k=1}^{n}frac{n}{k(2n-k+1)}-frac{1}{2}ln(n)$.

1

$begingroup$

Find $$lim_{nrightarrowinfty}int_{2npi}^{2(n+1)pi}xln xcos x,dx$$. Integrating by parts I obtained that the integral is equal to $$-int_{2npi}^{2(n+1)pi}ln x sin x, dx$$. Integrating again by parts, I get $$-int_{2npi}^{2(n+1)pi}ln xsin x,dx=lnfrac{2(n+1)pi}{2npi}-int_{2npi}^{2(n+1)pi}frac{cos x}{x}dx$$. I know that $$|frac{cos x}{x}|leq1$$ and by integrating it I get that $$int_{2npi}^{2(n+1)pi}frac{cos x}{x}in(-2pi,2pi)$$. But this does not provide me with a result. Any help?

calculus integration limits trigonometry definite-integrals

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edited 14 hours ago

MATHS MOD

647

asked 15 hours ago

user537002

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Within the integral, $2npi le x le 2(n+1)pi$, so $left|frac{cos x}{x}right| le frac{1}{2npi}$. Does that help?
  $endgroup$
  – FredH
  15 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Find $$lim_{nrightarrowinfty}int_{2npi}^{2(n+1)pi}xln xcos x,dx$$. Integrating by parts I obtained that the integral is equal to $$-int_{2npi}^{2(n+1)pi}ln x sin x, dx$$. Integrating again by parts, I get $$-int_{2npi}^{2(n+1)pi}ln xsin x,dx=lnfrac{2(n+1)pi}{2npi}-int_{2npi}^{2(n+1)pi}frac{cos x}{x}dx$$. I know that $$|frac{cos x}{x}|leq1$$ and by integrating it I get that $$int_{2npi}^{2(n+1)pi}frac{cos x}{x}in(-2pi,2pi)$$. But this does not provide me with a result. Any help?

calculus integration limits trigonometry definite-integrals

share|cite|improve this question

edited 14 hours ago

MATHS MOD

647

asked 15 hours ago

user537002

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Within the integral, $2npi le x le 2(n+1)pi$, so $left|frac{cos x}{x}right| le frac{1}{2npi}$. Does that help?
  $endgroup$
  – FredH
  15 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Find $$lim_{nrightarrowinfty}int_{2npi}^{2(n+1)pi}xln xcos x,dx$$. Integrating by parts I obtained that the integral is equal to $$-int_{2npi}^{2(n+1)pi}ln x sin x, dx$$. Integrating again by parts, I get $$-int_{2npi}^{2(n+1)pi}ln xsin x,dx=lnfrac{2(n+1)pi}{2npi}-int_{2npi}^{2(n+1)pi}frac{cos x}{x}dx$$. I know that $$|frac{cos x}{x}|leq1$$ and by integrating it I get that $$int_{2npi}^{2(n+1)pi}frac{cos x}{x}in(-2pi,2pi)$$. But this does not provide me with a result. Any help?

calculus integration limits trigonometry definite-integrals

share|cite|improve this question

edited 14 hours ago

MATHS MOD

647

asked 15 hours ago

user537002

$endgroup$

share|cite|improve this question

edited 14 hours ago

MATHS MOD

647

asked 15 hours ago

user537002

share|cite|improve this question

edited 14 hours ago

MATHS MOD

647

asked 15 hours ago

user537002

edited 14 hours ago

MATHS MOD

647

edited 14 hours ago

MATHS MOD

647

2 Answers
2

active

oldest

votes

1

$begingroup$

NOT A SOLUTION:
This may or may not be of help. I hope it's of use.

Here we will address your limit by fist addressing the integral:
begin{equation}
I = int_{2npi}^{2(n + 1)pi}x ln(x) cos(x):dxnonumber
end{equation}
We first observe that:
begin{equation}
lim_{a rightarrow 0^+}frac{partial}{partial a} x^a =ln(x)nonumber
end{equation}
Thus (and by the Dominated Convergence Theorem and Leibniz's Integral Rule $I$ becomes:
begin{equation}
I = int_{2npi}^{2(n + 1)pi}x left( lim_{a rightarrow 0^+}frac{partial}{partial a} x^a right) cos(x):dx = lim_{a rightarrow 0^+}frac{partial}{partial a} int_{2npi}^{2(n + 1)pi}x^{a + 1}cos(x):dxnonumber
end{equation}
Here we make the substitution $x = t + 2npi$:
begin{equation}
I = lim_{a rightarrow 0^+}frac{partial}{partial a} int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(t + 2npi right):dt
end{equation}
Now $cosleft(t + 2npi right) = cosleft(tright)$. Thus:
begin{equation}
I = lim_{a rightarrow 0^+}frac{partial}{partial a} int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(tright):dtnonumber
end{equation}

In terms of a closed form, I am unsure how to continue (if the upper limit of the integral was infinity it would be easy!)
Of we now incorporate the limit (I will call the result $J$):
begin{equation}
J = lim_{n rightarrow infty} lim_{a rightarrow 0^+}frac{partial}{partial a}int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(tright):dtnonumber
end{equation}

share|cite|improve this answer

edited 11 hours ago

answered 12 hours ago

DavidGDavidG

2,4961726

$endgroup$

add a comment | 

0

$begingroup$

You obtained $$lnfrac{2(n+1)pi}{2npi}-int_{2npi}^{2(n+1)pi}frac{cos x}{x}dx.$$

You are further along than you think. In the first expression, you are applying $ln$ to an expression that has limit $1.$ In the integral, you gave away the ranch. You wrote $|(cos x)/x |le 1$ but much more is true: $|(cos x)/x |le 1/x.$

share|cite|improve this answer

answered 1 hour ago

zhw.zhw.

73.9k43175

$endgroup$

add a comment | 

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1

$begingroup$

NOT A SOLUTION:
This may or may not be of help. I hope it's of use.

Here we will address your limit by fist addressing the integral:
begin{equation}
I = int_{2npi}^{2(n + 1)pi}x ln(x) cos(x):dxnonumber
end{equation}
We first observe that:
begin{equation}
lim_{a rightarrow 0^+}frac{partial}{partial a} x^a =ln(x)nonumber
end{equation}
Thus (and by the Dominated Convergence Theorem and Leibniz's Integral Rule $I$ becomes:
begin{equation}
I = int_{2npi}^{2(n + 1)pi}x left( lim_{a rightarrow 0^+}frac{partial}{partial a} x^a right) cos(x):dx = lim_{a rightarrow 0^+}frac{partial}{partial a} int_{2npi}^{2(n + 1)pi}x^{a + 1}cos(x):dxnonumber
end{equation}
Here we make the substitution $x = t + 2npi$:
begin{equation}
I = lim_{a rightarrow 0^+}frac{partial}{partial a} int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(t + 2npi right):dt
end{equation}
Now $cosleft(t + 2npi right) = cosleft(tright)$. Thus:
begin{equation}
I = lim_{a rightarrow 0^+}frac{partial}{partial a} int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(tright):dtnonumber
end{equation}

In terms of a closed form, I am unsure how to continue (if the upper limit of the integral was infinity it would be easy!)
Of we now incorporate the limit (I will call the result $J$):
begin{equation}
J = lim_{n rightarrow infty} lim_{a rightarrow 0^+}frac{partial}{partial a}int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(tright):dtnonumber
end{equation}

share|cite|improve this answer

edited 11 hours ago

answered 12 hours ago

DavidGDavidG

2,4961726

$endgroup$

add a comment | 

1

$begingroup$

NOT A SOLUTION:
This may or may not be of help. I hope it's of use.

Here we will address your limit by fist addressing the integral:
begin{equation}
I = int_{2npi}^{2(n + 1)pi}x ln(x) cos(x):dxnonumber
end{equation}
We first observe that:
begin{equation}
lim_{a rightarrow 0^+}frac{partial}{partial a} x^a =ln(x)nonumber
end{equation}
Thus (and by the Dominated Convergence Theorem and Leibniz's Integral Rule $I$ becomes:
begin{equation}
I = int_{2npi}^{2(n + 1)pi}x left( lim_{a rightarrow 0^+}frac{partial}{partial a} x^a right) cos(x):dx = lim_{a rightarrow 0^+}frac{partial}{partial a} int_{2npi}^{2(n + 1)pi}x^{a + 1}cos(x):dxnonumber
end{equation}
Here we make the substitution $x = t + 2npi$:
begin{equation}
I = lim_{a rightarrow 0^+}frac{partial}{partial a} int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(t + 2npi right):dt
end{equation}
Now $cosleft(t + 2npi right) = cosleft(tright)$. Thus:
begin{equation}
I = lim_{a rightarrow 0^+}frac{partial}{partial a} int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(tright):dtnonumber
end{equation}

In terms of a closed form, I am unsure how to continue (if the upper limit of the integral was infinity it would be easy!)
Of we now incorporate the limit (I will call the result $J$):
begin{equation}
J = lim_{n rightarrow infty} lim_{a rightarrow 0^+}frac{partial}{partial a}int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(tright):dtnonumber
end{equation}

share|cite|improve this answer

edited 11 hours ago

answered 12 hours ago

DavidGDavidG

2,4961726

$endgroup$

add a comment | 

$begingroup$

NOT A SOLUTION:
This may or may not be of help. I hope it's of use.

Here we will address your limit by fist addressing the integral:
begin{equation}
I = int_{2npi}^{2(n + 1)pi}x ln(x) cos(x):dxnonumber
end{equation}
We first observe that:
begin{equation}
lim_{a rightarrow 0^+}frac{partial}{partial a} x^a =ln(x)nonumber
end{equation}
Thus (and by the Dominated Convergence Theorem and Leibniz's Integral Rule $I$ becomes:
begin{equation}
I = int_{2npi}^{2(n + 1)pi}x left( lim_{a rightarrow 0^+}frac{partial}{partial a} x^a right) cos(x):dx = lim_{a rightarrow 0^+}frac{partial}{partial a} int_{2npi}^{2(n + 1)pi}x^{a + 1}cos(x):dxnonumber
end{equation}
Here we make the substitution $x = t + 2npi$:
begin{equation}
I = lim_{a rightarrow 0^+}frac{partial}{partial a} int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(t + 2npi right):dt
end{equation}
Now $cosleft(t + 2npi right) = cosleft(tright)$. Thus:
begin{equation}
I = lim_{a rightarrow 0^+}frac{partial}{partial a} int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(tright):dtnonumber
end{equation}

In terms of a closed form, I am unsure how to continue (if the upper limit of the integral was infinity it would be easy!)
Of we now incorporate the limit (I will call the result $J$):
begin{equation}
J = lim_{n rightarrow infty} lim_{a rightarrow 0^+}frac{partial}{partial a}int_0^{2pi} left(t + 2npiright)^{a + 1} cosleft(tright):dtnonumber
end{equation}

share|cite|improve this answer

edited 11 hours ago

answered 12 hours ago

DavidGDavidG

2,4961726

$endgroup$

share|cite|improve this answer

edited 11 hours ago

answered 12 hours ago

DavidGDavidG

2,4961726

answered 12 hours ago

DavidGDavidG

2,4961726

answered 12 hours ago

DavidGDavidG

2,4961726

0

$begingroup$

You obtained $$lnfrac{2(n+1)pi}{2npi}-int_{2npi}^{2(n+1)pi}frac{cos x}{x}dx.$$

You are further along than you think. In the first expression, you are applying $ln$ to an expression that has limit $1.$ In the integral, you gave away the ranch. You wrote $|(cos x)/x |le 1$ but much more is true: $|(cos x)/x |le 1/x.$

share|cite|improve this answer

answered 1 hour ago

zhw.zhw.

73.9k43175

$endgroup$

add a comment | 

0

$begingroup$

You obtained $$lnfrac{2(n+1)pi}{2npi}-int_{2npi}^{2(n+1)pi}frac{cos x}{x}dx.$$

You are further along than you think. In the first expression, you are applying $ln$ to an expression that has limit $1.$ In the integral, you gave away the ranch. You wrote $|(cos x)/x |le 1$ but much more is true: $|(cos x)/x |le 1/x.$

share|cite|improve this answer

answered 1 hour ago

zhw.zhw.

73.9k43175

$endgroup$

add a comment | 

$begingroup$

You obtained $$lnfrac{2(n+1)pi}{2npi}-int_{2npi}^{2(n+1)pi}frac{cos x}{x}dx.$$

You are further along than you think. In the first expression, you are applying $ln$ to an expression that has limit $1.$ In the integral, you gave away the ranch. You wrote $|(cos x)/x |le 1$ but much more is true: $|(cos x)/x |le 1/x.$

share|cite|improve this answer

answered 1 hour ago

zhw.zhw.

73.9k43175

$endgroup$

share|cite|improve this answer

answered 1 hour ago

zhw.zhw.

73.9k43175

answered 1 hour ago

zhw.zhw.

73.9k43175

answered 1 hour ago

zhw.zhw.

73.9k43175