Prove that $int_x^x + ln xover x e^t^2 ,dt> e^x^2ln xover2x$If $mu(X)=1$, then $lim_prightarrow 0|f|_p=expleft(int_Xln(|f|)dmuright)$Gaussian integrals over a half-spaceDifficult integral: $int_0^ln2 sqrte^2x -2 +e^-2xover e^x+e^-x,dx$Multiple Integration order doesn't agree.Solve $ int_0^sqrtpi / 2left(int_x^sqrtpi / 2 sin(y^2) dy right)dx$Double Integral over regionEvaluate $int_0^1int_x^1 e^x/y dy,dx$Estimates of $f(x) = int_x^x^2 dfracdyln y$Help with an Incomplete Gamma function-like integral $int_x^infty t^m-1e^-0.5 t^2 pm pt dt$$exists c>0$ such that $ (z-x)int_z^yf(t)dt - (y-z)int_x^zf(t)dt geq c(z-x)(y-z)$
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Prove that $int_x^x + ln xover x e^t^2 ,dt> e^x^2ln xover2x$
If $mu(X)=1$, then $lim_prightarrow 0|f|_p=expleft(int_Xln(|f|)dmuright)$Gaussian integrals over a half-spaceDifficult integral: $int_0^ln2 sqrte^2x -2 +e^-2xover e^x+e^-x,dx$Multiple Integration order doesn't agree.Solve $ int_0^sqrtpi / 2left(int_x^sqrtpi / 2 sin(y^2) dy right)dx$Double Integral over regionEvaluate $int_0^1int_x^1 e^x/y dy,dx$Estimates of $f(x) = int_x^x^2 dfracdyln y$Help with an Incomplete Gamma function-like integral $int_x^infty t^m-1e^-0.5 t^2 pm pt dt$$exists c>0$ such that $ (z-x)int_z^yf(t)dt - (y-z)int_x^zf(t)dt geq c(z-x)(y-z)$
$begingroup$
$$int_x^x + ln xover x e^t^2 ,dt> e^x^2ln xover2x\x > 1,$$
I wanted to start with integrating $e^t^2$, but unfortunately it's not helpful. What's the way to solve inequality like that?
integration definite-integrals
edited Mar 29 at 22:03
Bernard
124k741117
asked Mar 29 at 21:55
MichaelMichael
266
$endgroup$
add a comment |
$begingroup$
$$int_x^x + ln xover x e^t^2 ,dt> e^x^2ln xover2x\x > 1,$$
I wanted to start with integrating $e^t^2$, but unfortunately it's not helpful. What's the way to solve inequality like that?
integration definite-integrals
edited Mar 29 at 22:03
Bernard
124k741117
asked Mar 29 at 21:55
MichaelMichael
266
$endgroup$
add a comment |
$begingroup$
$$int_x^x + ln xover x e^t^2 ,dt> e^x^2ln xover2x\x > 1,$$
I wanted to start with integrating $e^t^2$, but unfortunately it's not helpful. What's the way to solve inequality like that?
integration definite-integrals
edited Mar 29 at 22:03
Bernard
124k741117
asked Mar 29 at 21:55
MichaelMichael
266
$endgroup$
$$int_x^x + ln xover x e^t^2 ,dt> e^x^2ln xover2x\x > 1,$$
I wanted to start with integrating $e^t^2$, but unfortunately it's not helpful. What's the way to solve inequality like that?
integration definite-integrals
integration definite-integrals
edited Mar 29 at 22:03
Bernard
124k741117
asked Mar 29 at 21:55
MichaelMichael
266
edited Mar 29 at 22:03
Bernard
124k741117
asked Mar 29 at 21:55
MichaelMichael
266
edited Mar 29 at 22:03
Bernard
124k741117
edited Mar 29 at 22:03
Bernard
124k741117
edited Mar 29 at 22:03
Bernard
124k741117
124k741117
asked Mar 29 at 21:55
MichaelMichael
266
asked Mar 29 at 21:55
MichaelMichael
266
asked Mar 29 at 21:55
MichaelMichael
266
266
add a comment |
add a comment |
1 Answer
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$begingroup$
Any integral which has a positive increasing integrand is greater than or equal to its minimum value multiplied by the width of the integration region. So, in this case
$$int_x^x + lnxover x e^t^2 ,dtgefrace^x^2ln(x)xgtfrace^x^2ln(x)2x$$
answered Mar 29 at 22:05
Peter ForemanPeter Foreman
6,2261317
$endgroup$
$begingroup$
And by looking at the 2nd derivative, you can get better estimates. I feel that OP was missing something because this answer was so immediate.
$endgroup$
– marty cohen
Mar 29 at 22:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Any integral which has a positive increasing integrand is greater than or equal to its minimum value multiplied by the width of the integration region. So, in this case
$$int_x^x + lnxover x e^t^2 ,dtgefrace^x^2ln(x)xgtfrace^x^2ln(x)2x$$
answered Mar 29 at 22:05
Peter ForemanPeter Foreman
6,2261317
$endgroup$
$begingroup$
And by looking at the 2nd derivative, you can get better estimates. I feel that OP was missing something because this answer was so immediate.
$endgroup$
– marty cohen
Mar 29 at 22:46
add a comment |
$begingroup$
Any integral which has a positive increasing integrand is greater than or equal to its minimum value multiplied by the width of the integration region. So, in this case
$$int_x^x + lnxover x e^t^2 ,dtgefrace^x^2ln(x)xgtfrace^x^2ln(x)2x$$
answered Mar 29 at 22:05
Peter ForemanPeter Foreman
6,2261317
$endgroup$
$begingroup$
And by looking at the 2nd derivative, you can get better estimates. I feel that OP was missing something because this answer was so immediate.
$endgroup$
– marty cohen
Mar 29 at 22:46
add a comment |
$begingroup$
Any integral which has a positive increasing integrand is greater than or equal to its minimum value multiplied by the width of the integration region. So, in this case
$$int_x^x + lnxover x e^t^2 ,dtgefrace^x^2ln(x)xgtfrace^x^2ln(x)2x$$
answered Mar 29 at 22:05
Peter ForemanPeter Foreman
6,2261317
$endgroup$
Any integral which has a positive increasing integrand is greater than or equal to its minimum value multiplied by the width of the integration region. So, in this case
$$int_x^x + lnxover x e^t^2 ,dtgefrace^x^2ln(x)xgtfrace^x^2ln(x)2x$$
answered Mar 29 at 22:05
Peter ForemanPeter Foreman
6,2261317
answered Mar 29 at 22:05
Peter ForemanPeter Foreman
6,2261317
answered Mar 29 at 22:05
Peter ForemanPeter Foreman
6,2261317
answered Mar 29 at 22:05
Peter ForemanPeter Foreman
6,2261317
6,2261317
$begingroup$
And by looking at the 2nd derivative, you can get better estimates. I feel that OP was missing something because this answer was so immediate.
$endgroup$
– marty cohen
Mar 29 at 22:46
add a comment |
$begingroup$
And by looking at the 2nd derivative, you can get better estimates. I feel that OP was missing something because this answer was so immediate.
$endgroup$
– marty cohen
Mar 29 at 22:46
$begingroup$
And by looking at the 2nd derivative, you can get better estimates. I feel that OP was missing something because this answer was so immediate.
$endgroup$
– marty cohen
Mar 29 at 22:46
$begingroup$
And by looking at the 2nd derivative, you can get better estimates. I feel that OP was missing something because this answer was so immediate.
$endgroup$
– marty cohen
Mar 29 at 22:46
add a comment |
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