Limit and integral with trigonometric function
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Calculate $$lim_{n rightarrow infty} int_{2npi}^{2(n+1)pi} xln(x)cos xmathrm{d}x.$$
Firstly, I tried to solve it by parts, but I think this is not a good option. Then, I tried to use the average theorem, but $cos x$ isn't positive for the whole interval and the other options don't help me too much, I guess. Please help me!
integration limits
edited Dec 22 '18 at 16:05
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked Dec 22 '18 at 15:46
Ababei AbabeiAbabei Ababei
62
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add a comment |
$begingroup$
Calculate $$lim_{n rightarrow infty} int_{2npi}^{2(n+1)pi} xln(x)cos xmathrm{d}x.$$
Firstly, I tried to solve it by parts, but I think this is not a good option. Then, I tried to use the average theorem, but $cos x$ isn't positive for the whole interval and the other options don't help me too much, I guess. Please help me!
integration limits
edited Dec 22 '18 at 16:05
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked Dec 22 '18 at 15:46
Ababei AbabeiAbabei Ababei
62
$endgroup$
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Hint. Riemann-Lebesgue Lemma.
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– Von Neumann
Dec 22 '18 at 16:31
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@VonNeumann How does RLL help?
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– zhw.
Dec 22 '18 at 18:37
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Hint: Integrate by parts twice.
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– zhw.
Dec 22 '18 at 18:41
add a comment |
$begingroup$
Calculate $$lim_{n rightarrow infty} int_{2npi}^{2(n+1)pi} xln(x)cos xmathrm{d}x.$$
Firstly, I tried to solve it by parts, but I think this is not a good option. Then, I tried to use the average theorem, but $cos x$ isn't positive for the whole interval and the other options don't help me too much, I guess. Please help me!
integration limits
edited Dec 22 '18 at 16:05
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked Dec 22 '18 at 15:46
Ababei AbabeiAbabei Ababei
62
$endgroup$
Calculate $$lim_{n rightarrow infty} int_{2npi}^{2(n+1)pi} xln(x)cos xmathrm{d}x.$$
Firstly, I tried to solve it by parts, but I think this is not a good option. Then, I tried to use the average theorem, but $cos x$ isn't positive for the whole interval and the other options don't help me too much, I guess. Please help me!
integration limits
integration limits
edited Dec 22 '18 at 16:05
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked Dec 22 '18 at 15:46
Ababei AbabeiAbabei Ababei
62
edited Dec 22 '18 at 16:05
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked Dec 22 '18 at 15:46
Ababei AbabeiAbabei Ababei
62
edited Dec 22 '18 at 16:05
GNUSupporter 8964民主女神 地下教會
14.1k82651
edited Dec 22 '18 at 16:05
GNUSupporter 8964民主女神 地下教會
14.1k82651
edited Dec 22 '18 at 16:05
GNUSupporter 8964民主女神 地下教會
14.1k82651
14.1k82651
asked Dec 22 '18 at 15:46
Ababei AbabeiAbabei Ababei
62
asked Dec 22 '18 at 15:46
Ababei AbabeiAbabei Ababei
62
asked Dec 22 '18 at 15:46
Ababei AbabeiAbabei Ababei
62
62
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Hint. Riemann-Lebesgue Lemma.
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– Von Neumann
Dec 22 '18 at 16:31
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@VonNeumann How does RLL help?
$endgroup$
– zhw.
Dec 22 '18 at 18:37
$begingroup$
Hint: Integrate by parts twice.
$endgroup$
– zhw.
Dec 22 '18 at 18:41
add a comment |
$begingroup$
Hint. Riemann-Lebesgue Lemma.
$endgroup$
– Von Neumann
Dec 22 '18 at 16:31
$begingroup$
@VonNeumann How does RLL help?
$endgroup$
– zhw.
Dec 22 '18 at 18:37
$begingroup$
Hint: Integrate by parts twice.
$endgroup$
– zhw.
Dec 22 '18 at 18:41
$begingroup$
Hint. Riemann-Lebesgue Lemma.
$endgroup$
– Von Neumann
Dec 22 '18 at 16:31
$begingroup$
Hint. Riemann-Lebesgue Lemma.
$endgroup$
– Von Neumann
Dec 22 '18 at 16:31
$begingroup$
@VonNeumann How does RLL help?
$endgroup$
– zhw.
Dec 22 '18 at 18:37
$begingroup$
@VonNeumann How does RLL help?
$endgroup$
– zhw.
Dec 22 '18 at 18:37
$begingroup$
Hint: Integrate by parts twice.
$endgroup$
– zhw.
Dec 22 '18 at 18:41
$begingroup$
Hint: Integrate by parts twice.
$endgroup$
– zhw.
Dec 22 '18 at 18:41
add a comment |
1 Answer
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You need to integrate by parts twice. $int_{2npi}^{(2n+1)pi} xln(x)cosxdx=xln(x)sinxbig]_{2npi}^{(2n+1)pi}-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx$. Since $sin(npi)=0$, the expression is $-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx=cosx(1+ln(x))big]_{2npi}^{(2n+1)pi}+int_{2npi}^{(2n+1)pi}frac{cosx}{x}dx$.
The last integral $to 0$ as $nto infty$, so the original integral ends up as $-(2+ln(2npi)+ln((2n+1)pi)to -infty$.
Note this is not too surprising since the term $xln(x)$ is larger in the second half of the integration interval, where the $cosx$ is negative.
answered Dec 22 '18 at 18:42
herb steinbergherb steinberg
3,1982311
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1 Answer
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1 Answer
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$begingroup$
You need to integrate by parts twice. $int_{2npi}^{(2n+1)pi} xln(x)cosxdx=xln(x)sinxbig]_{2npi}^{(2n+1)pi}-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx$. Since $sin(npi)=0$, the expression is $-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx=cosx(1+ln(x))big]_{2npi}^{(2n+1)pi}+int_{2npi}^{(2n+1)pi}frac{cosx}{x}dx$.
The last integral $to 0$ as $nto infty$, so the original integral ends up as $-(2+ln(2npi)+ln((2n+1)pi)to -infty$.
Note this is not too surprising since the term $xln(x)$ is larger in the second half of the integration interval, where the $cosx$ is negative.
answered Dec 22 '18 at 18:42
herb steinbergherb steinberg
3,1982311
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add a comment |
$begingroup$
You need to integrate by parts twice. $int_{2npi}^{(2n+1)pi} xln(x)cosxdx=xln(x)sinxbig]_{2npi}^{(2n+1)pi}-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx$. Since $sin(npi)=0$, the expression is $-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx=cosx(1+ln(x))big]_{2npi}^{(2n+1)pi}+int_{2npi}^{(2n+1)pi}frac{cosx}{x}dx$.
The last integral $to 0$ as $nto infty$, so the original integral ends up as $-(2+ln(2npi)+ln((2n+1)pi)to -infty$.
Note this is not too surprising since the term $xln(x)$ is larger in the second half of the integration interval, where the $cosx$ is negative.
answered Dec 22 '18 at 18:42
herb steinbergherb steinberg
3,1982311
$endgroup$
add a comment |
$begingroup$
You need to integrate by parts twice. $int_{2npi}^{(2n+1)pi} xln(x)cosxdx=xln(x)sinxbig]_{2npi}^{(2n+1)pi}-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx$. Since $sin(npi)=0$, the expression is $-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx=cosx(1+ln(x))big]_{2npi}^{(2n+1)pi}+int_{2npi}^{(2n+1)pi}frac{cosx}{x}dx$.
The last integral $to 0$ as $nto infty$, so the original integral ends up as $-(2+ln(2npi)+ln((2n+1)pi)to -infty$.
Note this is not too surprising since the term $xln(x)$ is larger in the second half of the integration interval, where the $cosx$ is negative.
answered Dec 22 '18 at 18:42
herb steinbergherb steinberg
3,1982311
$endgroup$
You need to integrate by parts twice. $int_{2npi}^{(2n+1)pi} xln(x)cosxdx=xln(x)sinxbig]_{2npi}^{(2n+1)pi}-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx$. Since $sin(npi)=0$, the expression is $-int_{2npi}^{(2n+1)pi}(1+ln(x))sinxdx=cosx(1+ln(x))big]_{2npi}^{(2n+1)pi}+int_{2npi}^{(2n+1)pi}frac{cosx}{x}dx$.
The last integral $to 0$ as $nto infty$, so the original integral ends up as $-(2+ln(2npi)+ln((2n+1)pi)to -infty$.
Note this is not too surprising since the term $xln(x)$ is larger in the second half of the integration interval, where the $cosx$ is negative.
answered Dec 22 '18 at 18:42
herb steinbergherb steinberg
3,1982311
answered Dec 22 '18 at 18:42
herb steinbergherb steinberg
3,1982311
answered Dec 22 '18 at 18:42
herb steinbergherb steinberg
3,1982311
answered Dec 22 '18 at 18:42
herb steinbergherb steinberg
3,1982311
3,1982311
add a comment |
add a comment |
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$begingroup$
Hint. Riemann-Lebesgue Lemma.
$endgroup$
– Von Neumann
Dec 22 '18 at 16:31
$begingroup$
@VonNeumann How does RLL help?
$endgroup$
– zhw.
Dec 22 '18 at 18:37
$begingroup$
Hint: Integrate by parts twice.
$endgroup$
– zhw.
Dec 22 '18 at 18:41