Are functions like $frac{2cos(x-frac{1}{x})}{x^2+2}$ integrable in any way?
0
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Is there a way to integrate $$frac{2cos(x-frac{1}{x})}{x^2+2},$$ perhaps through methods other than Riemann integration? I was trying to use Glasser's master theorem to integrate $$frac{cos(x)}{x^4+1},$$ but when I graph the former function (after substitution this is the function), it does not look possible. Glasser's master theorem states that both functions must be integrable for the substitution to occur, so I wonder if it is possible .
real-analysis integration definite-integrals continuity
edited Dec 14 '18 at 19:09
amWhy
1
asked Dec 14 '18 at 19:01
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
add a comment |
0
$begingroup$
Is there a way to integrate $$frac{2cos(x-frac{1}{x})}{x^2+2},$$ perhaps through methods other than Riemann integration? I was trying to use Glasser's master theorem to integrate $$frac{cos(x)}{x^4+1},$$ but when I graph the former function (after substitution this is the function), it does not look possible. Glasser's master theorem states that both functions must be integrable for the substitution to occur, so I wonder if it is possible .
real-analysis integration definite-integrals continuity
edited Dec 14 '18 at 19:09
amWhy
1
asked Dec 14 '18 at 19:01
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
1
$begingroup$
Glasser's master theorem applies to principal value integral over the whole real line. Is that what you are interested in?
$endgroup$
– Fabian
Dec 14 '18 at 20:10
$begingroup$
@Fabian yes that's what I'm looking for
$endgroup$
– Suchetan Dontha
Dec 14 '18 at 20:12
$begingroup$
I didn't know this theorem before. But Wikipedia says that "It is applicable in cases where the integrals must be construed as Cauchy principal values, and a fortiori it is applicable when the integral converges absolutely." But we clearly have $int |2cos (x -x^{-1})/(2+x^2)|dx leq 2int 1/(2+x^2)dx <infty$, so the theorem is applicable. This does not tell you, however, how you compute $int cos(x)/(1+x^4)dx$.
$endgroup$
– PhoemueX
Dec 14 '18 at 20:32
$begingroup$
Glasser's Master Theorem can certainly be applied to the numerator, in the current form you can not apply it to the denominator and thus the full integrand. Not to say it can't be done with some rearranging though.
$endgroup$
– DavidG
Dec 16 '18 at 0:47
$begingroup$
How did you go from $cos(x)/(x^4 + 1)$ to $2cos(x - 1/x)/(x^2 + 2)$ ??
$endgroup$
– DavidG
Dec 16 '18 at 0:51
add a comment |
0
0
0
$begingroup$
Is there a way to integrate $$frac{2cos(x-frac{1}{x})}{x^2+2},$$ perhaps through methods other than Riemann integration? I was trying to use Glasser's master theorem to integrate $$frac{cos(x)}{x^4+1},$$ but when I graph the former function (after substitution this is the function), it does not look possible. Glasser's master theorem states that both functions must be integrable for the substitution to occur, so I wonder if it is possible .
real-analysis integration definite-integrals continuity
edited Dec 14 '18 at 19:09
amWhy
1
asked Dec 14 '18 at 19:01
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
Is there a way to integrate $$frac{2cos(x-frac{1}{x})}{x^2+2},$$ perhaps through methods other than Riemann integration? I was trying to use Glasser's master theorem to integrate $$frac{cos(x)}{x^4+1},$$ but when I graph the former function (after substitution this is the function), it does not look possible. Glasser's master theorem states that both functions must be integrable for the substitution to occur, so I wonder if it is possible .
real-analysis integration definite-integrals continuity
real-analysis integration definite-integrals continuity
edited Dec 14 '18 at 19:09
amWhy
1
asked Dec 14 '18 at 19:01
Suchetan DonthaSuchetan Dontha
14312
edited Dec 14 '18 at 19:09
amWhy
1
asked Dec 14 '18 at 19:01
Suchetan DonthaSuchetan Dontha
14312
edited Dec 14 '18 at 19:09
amWhy
1
edited Dec 14 '18 at 19:09
amWhy
1
edited Dec 14 '18 at 19:09
amWhy
1
1
asked Dec 14 '18 at 19:01
Suchetan DonthaSuchetan Dontha
14312
asked Dec 14 '18 at 19:01
Suchetan DonthaSuchetan Dontha
14312
asked Dec 14 '18 at 19:01
Suchetan DonthaSuchetan Dontha
14312
14312
1
$begingroup$
Glasser's master theorem applies to principal value integral over the whole real line. Is that what you are interested in?
$endgroup$
– Fabian
Dec 14 '18 at 20:10
$begingroup$
@Fabian yes that's what I'm looking for
$endgroup$
– Suchetan Dontha
Dec 14 '18 at 20:12
$begingroup$
I didn't know this theorem before. But Wikipedia says that "It is applicable in cases where the integrals must be construed as Cauchy principal values, and a fortiori it is applicable when the integral converges absolutely." But we clearly have $int |2cos (x -x^{-1})/(2+x^2)|dx leq 2int 1/(2+x^2)dx <infty$, so the theorem is applicable. This does not tell you, however, how you compute $int cos(x)/(1+x^4)dx$.
$endgroup$
– PhoemueX
Dec 14 '18 at 20:32
$begingroup$
Glasser's Master Theorem can certainly be applied to the numerator, in the current form you can not apply it to the denominator and thus the full integrand. Not to say it can't be done with some rearranging though.
$endgroup$
– DavidG
Dec 16 '18 at 0:47
$begingroup$
How did you go from $cos(x)/(x^4 + 1)$ to $2cos(x - 1/x)/(x^2 + 2)$ ??
$endgroup$
– DavidG
Dec 16 '18 at 0:51
add a comment |
1
$begingroup$
Glasser's master theorem applies to principal value integral over the whole real line. Is that what you are interested in?
$endgroup$
– Fabian
Dec 14 '18 at 20:10
$begingroup$
@Fabian yes that's what I'm looking for
$endgroup$
– Suchetan Dontha
Dec 14 '18 at 20:12
$begingroup$
I didn't know this theorem before. But Wikipedia says that "It is applicable in cases where the integrals must be construed as Cauchy principal values, and a fortiori it is applicable when the integral converges absolutely." But we clearly have $int |2cos (x -x^{-1})/(2+x^2)|dx leq 2int 1/(2+x^2)dx <infty$, so the theorem is applicable. This does not tell you, however, how you compute $int cos(x)/(1+x^4)dx$.
$endgroup$
– PhoemueX
Dec 14 '18 at 20:32
$begingroup$
Glasser's Master Theorem can certainly be applied to the numerator, in the current form you can not apply it to the denominator and thus the full integrand. Not to say it can't be done with some rearranging though.
$endgroup$
– DavidG
Dec 16 '18 at 0:47
$begingroup$
How did you go from $cos(x)/(x^4 + 1)$ to $2cos(x - 1/x)/(x^2 + 2)$ ??
$endgroup$
– DavidG
Dec 16 '18 at 0:51
1
1
$begingroup$
Glasser's master theorem applies to principal value integral over the whole real line. Is that what you are interested in?
$endgroup$
– Fabian
Dec 14 '18 at 20:10
$begingroup$
Glasser's master theorem applies to principal value integral over the whole real line. Is that what you are interested in?
$endgroup$
– Fabian
Dec 14 '18 at 20:10
$begingroup$
@Fabian yes that's what I'm looking for
$endgroup$
– Suchetan Dontha
Dec 14 '18 at 20:12
$begingroup$
@Fabian yes that's what I'm looking for
$endgroup$
– Suchetan Dontha
Dec 14 '18 at 20:12
$begingroup$
I didn't know this theorem before. But Wikipedia says that "It is applicable in cases where the integrals must be construed as Cauchy principal values, and a fortiori it is applicable when the integral converges absolutely." But we clearly have $int |2cos (x -x^{-1})/(2+x^2)|dx leq 2int 1/(2+x^2)dx <infty$, so the theorem is applicable. This does not tell you, however, how you compute $int cos(x)/(1+x^4)dx$.
$endgroup$
– PhoemueX
Dec 14 '18 at 20:32
$begingroup$
I didn't know this theorem before. But Wikipedia says that "It is applicable in cases where the integrals must be construed as Cauchy principal values, and a fortiori it is applicable when the integral converges absolutely." But we clearly have $int |2cos (x -x^{-1})/(2+x^2)|dx leq 2int 1/(2+x^2)dx <infty$, so the theorem is applicable. This does not tell you, however, how you compute $int cos(x)/(1+x^4)dx$.
$endgroup$
– PhoemueX
Dec 14 '18 at 20:32
$begingroup$
Glasser's Master Theorem can certainly be applied to the numerator, in the current form you can not apply it to the denominator and thus the full integrand. Not to say it can't be done with some rearranging though.
$endgroup$
– DavidG
Dec 16 '18 at 0:47
$begingroup$
Glasser's Master Theorem can certainly be applied to the numerator, in the current form you can not apply it to the denominator and thus the full integrand. Not to say it can't be done with some rearranging though.
$endgroup$
– DavidG
Dec 16 '18 at 0:47
$begingroup$
How did you go from $cos(x)/(x^4 + 1)$ to $2cos(x - 1/x)/(x^2 + 2)$ ??
$endgroup$
– DavidG
Dec 16 '18 at 0:51
$begingroup$
How did you go from $cos(x)/(x^4 + 1)$ to $2cos(x - 1/x)/(x^2 + 2)$ ??
$endgroup$
– DavidG
Dec 16 '18 at 0:51
add a comment |
0
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1
$begingroup$
Glasser's master theorem applies to principal value integral over the whole real line. Is that what you are interested in?
$endgroup$
– Fabian
Dec 14 '18 at 20:10
$begingroup$
@Fabian yes that's what I'm looking for
$endgroup$
– Suchetan Dontha
Dec 14 '18 at 20:12
$begingroup$
I didn't know this theorem before. But Wikipedia says that "It is applicable in cases where the integrals must be construed as Cauchy principal values, and a fortiori it is applicable when the integral converges absolutely." But we clearly have $int |2cos (x -x^{-1})/(2+x^2)|dx leq 2int 1/(2+x^2)dx <infty$, so the theorem is applicable. This does not tell you, however, how you compute $int cos(x)/(1+x^4)dx$.
$endgroup$
– PhoemueX
Dec 14 '18 at 20:32
$begingroup$
Glasser's Master Theorem can certainly be applied to the numerator, in the current form you can not apply it to the denominator and thus the full integrand. Not to say it can't be done with some rearranging though.
$endgroup$
– DavidG
Dec 16 '18 at 0:47
$begingroup$
How did you go from $cos(x)/(x^4 + 1)$ to $2cos(x - 1/x)/(x^2 + 2)$ ??
$endgroup$
– DavidG
Dec 16 '18 at 0:51