Calculate $lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2$Solving $int_0^infty sin(ax^2)sin(b/x^2),mathrm dx$Is $int_0^infty left (int_0^infty f(k) k sin kr , mathrm dk right) mathrm dr = int_0^infty f(k) , mathrm dk$ correct?calculate $lim_ntoinftyint_[0,infty) exp(-x)sin(nx),mathrmdmathcalL^1(x)$Evaluating $int_0^infty fracxi x^alpha e^x-xi :mathrmdx$Computing $lim_epsilon rightarrow 0 int_0^infty fracsin xx arctanfracxepsilondx$Evaluating $lim_nrightarrow inftyint_0^infty frace^sin(ln(x))1+sqrt nx^42dx $Finding $lim_ntoinftyint_0^inftyfracx^n-21+x^n|sin(nx)|$.Putting limit inside an integral to calculate $int_0^infty fracsin xxdx = fracpi2$Computing $limlimits_n to infty int_0^fracpi2frac(sin(x))^n1-sin(x),mathrmdx $$lim_arightarrow 0 int_0^1 frac11+asin(x) , mathrmdx$
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Calculate $lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2$
Solving $int_0^infty sin(ax^2)sin(b/x^2),mathrm dx$Is $int_0^infty left (int_0^infty f(k) k sin kr , mathrm dk right) mathrm dr = int_0^infty f(k) , mathrm dk$ correct?calculate $lim_ntoinftyint_[0,infty) exp(-x)sin(nx),mathrmdmathcalL^1(x)$Evaluating $int_0^infty fracxi x^alpha e^x-xi :mathrmdx$Computing $lim_epsilon rightarrow 0 int_0^infty fracsin xx arctanfracxepsilondx$Evaluating $lim_nrightarrow inftyint_0^infty frace^sin(ln(x))1+sqrt nx^42dx $Finding $lim_ntoinftyint_0^inftyfracx^n-21+x^n|sin(nx)|$.Putting limit inside an integral to calculate $int_0^infty fracsin xxdx = fracpi2$Computing $limlimits_n to infty int_0^fracpi2frac(sin(x))^n1-sin(x),mathrmdx $$lim_arightarrow 0 int_0^1 frac11+asin(x) , mathrmdx$
$begingroup$
Calculate $$lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2$$, well first of all the function isn't monotonic, so I cant use the Lebesgue therem to go with the limit under the integral. What I could do is to express $sinx=x+O(x^3)$ then $$lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2=lim_ntoinftyint_0^nfracmathrmdxn+x=lim_ntoinfty ln2n-ln n=ln2$$, but i am not sure if this is correct and also would like to know why this is correct, as of course I ate the $O(x^3)$ when approximating sin function.
real-analysis calculus integration analysis improper-integrals
edited Mar 20 at 17:58
Rebellos
15.6k31250
asked Mar 20 at 17:54
ryszard egginkryszard eggink
415110
$endgroup$
add a comment |
$begingroup$
Calculate $$lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2$$, well first of all the function isn't monotonic, so I cant use the Lebesgue therem to go with the limit under the integral. What I could do is to express $sinx=x+O(x^3)$ then $$lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2=lim_ntoinftyint_0^nfracmathrmdxn+x=lim_ntoinfty ln2n-ln n=ln2$$, but i am not sure if this is correct and also would like to know why this is correct, as of course I ate the $O(x^3)$ when approximating sin function.
real-analysis calculus integration analysis improper-integrals
edited Mar 20 at 17:58
Rebellos
15.6k31250
asked Mar 20 at 17:54
ryszard egginkryszard eggink
415110
$endgroup$
4
$begingroup$
Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral.
$endgroup$
– Sangchul Lee
Mar 20 at 18:00
$begingroup$
nice suggestion!
$endgroup$
– ryszard eggink
Mar 20 at 18:02
add a comment |
$begingroup$
Calculate $$lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2$$, well first of all the function isn't monotonic, so I cant use the Lebesgue therem to go with the limit under the integral. What I could do is to express $sinx=x+O(x^3)$ then $$lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2=lim_ntoinftyint_0^nfracmathrmdxn+x=lim_ntoinfty ln2n-ln n=ln2$$, but i am not sure if this is correct and also would like to know why this is correct, as of course I ate the $O(x^3)$ when approximating sin function.
real-analysis calculus integration analysis improper-integrals
edited Mar 20 at 17:58
Rebellos
15.6k31250
asked Mar 20 at 17:54
ryszard egginkryszard eggink
415110
$endgroup$
Calculate $$lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2$$, well first of all the function isn't monotonic, so I cant use the Lebesgue therem to go with the limit under the integral. What I could do is to express $sinx=x+O(x^3)$ then $$lim_ntoinftyint_0^nfracmathrmdxn+n^2sinfracxn^2=lim_ntoinftyint_0^nfracmathrmdxn+x=lim_ntoinfty ln2n-ln n=ln2$$, but i am not sure if this is correct and also would like to know why this is correct, as of course I ate the $O(x^3)$ when approximating sin function.
real-analysis calculus integration analysis improper-integrals
real-analysis calculus integration analysis improper-integrals
edited Mar 20 at 17:58
Rebellos
15.6k31250
asked Mar 20 at 17:54
ryszard egginkryszard eggink
415110
edited Mar 20 at 17:58
Rebellos
15.6k31250
asked Mar 20 at 17:54
ryszard egginkryszard eggink
415110
edited Mar 20 at 17:58
Rebellos
15.6k31250
edited Mar 20 at 17:58
Rebellos
15.6k31250
edited Mar 20 at 17:58
Rebellos
15.6k31250
15.6k31250
asked Mar 20 at 17:54
ryszard egginkryszard eggink
415110
asked Mar 20 at 17:54
ryszard egginkryszard eggink
415110
asked Mar 20 at 17:54
ryszard egginkryszard eggink
415110
415110
4
$begingroup$
Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral.
$endgroup$
– Sangchul Lee
Mar 20 at 18:00
$begingroup$
nice suggestion!
$endgroup$
– ryszard eggink
Mar 20 at 18:02
add a comment |
4
$begingroup$
Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral.
$endgroup$
– Sangchul Lee
Mar 20 at 18:00
$begingroup$
nice suggestion!
$endgroup$
– ryszard eggink
Mar 20 at 18:02
4
4
$begingroup$
Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral.
$endgroup$
– Sangchul Lee
Mar 20 at 18:00
$begingroup$
Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral.
$endgroup$
– Sangchul Lee
Mar 20 at 18:00
$begingroup$
nice suggestion!
$endgroup$
– ryszard eggink
Mar 20 at 18:02
$begingroup$
nice suggestion!
$endgroup$
– ryszard eggink
Mar 20 at 18:02
add a comment |
1 Answer
1
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$begingroup$
Hint
Since $u-u^3over 6le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $$1over n+xlefrac1n+n^2sinfracxn^2le frac1n+x-x^3over n^4le1over n+x-1over n$$and integrate the sides.
answered Mar 20 at 18:13
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint
Since $u-u^3over 6le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $$1over n+xlefrac1n+n^2sinfracxn^2le frac1n+x-x^3over n^4le1over n+x-1over n$$and integrate the sides.
answered Mar 20 at 18:13
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
Hint
Since $u-u^3over 6le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $$1over n+xlefrac1n+n^2sinfracxn^2le frac1n+x-x^3over n^4le1over n+x-1over n$$and integrate the sides.
answered Mar 20 at 18:13
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
Hint
Since $u-u^3over 6le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $$1over n+xlefrac1n+n^2sinfracxn^2le frac1n+x-x^3over n^4le1over n+x-1over n$$and integrate the sides.
answered Mar 20 at 18:13
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
Hint
Since $u-u^3over 6le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $$1over n+xlefrac1n+n^2sinfracxn^2le frac1n+x-x^3over n^4le1over n+x-1over n$$and integrate the sides.
answered Mar 20 at 18:13
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 20 at 18:13
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 20 at 18:13
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 20 at 18:13
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
add a comment |
add a comment |
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$begingroup$
Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral.
$endgroup$
– Sangchul Lee
Mar 20 at 18:00
$begingroup$
nice suggestion!
$endgroup$
– ryszard eggink
Mar 20 at 18:02