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Trying to prove an integral inequality
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Let $f_s,f_l:(0,infty) mapsto {mathbb R}$ be monotonic increasing functions with $f_s(x) leq f_l(x)$ and let $x_s,x_l$ be the smallest positive roots of the equation $1-f_{s,l}(x)=0$ respectively (assuming they exist, are real and no double roots). Then I believe the following is true
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} geq int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}$$
and if it is true I would like to proof it.
I just don‘t know where to start.
Heuristically I think it is because the increase in integration interval $x_s geq x_l$ outweights the decrease of the integrand values $$frac{{rm d}x}{sqrt{1-f_s(x)}} leq frac{{rm d}x}{sqrt{1-f_l(x)}} , .$$
Observe for small deviations $f_l = f_s + delta f$ we have to first order for the RHS
$$
int_0^{x_l} frac{{rm d}x}{sqrt{1-f_s(x)}} left( 1 + frac{delta f(x)}{2(1-f_s(x))} right)
$$
and so this requires
$$
int_{x_l}^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} geq int_0^{x_l} frac{{rm d}x}{sqrt{1-f_s(x)}} left( frac{delta f(x)}{2(1-f_s(x))} right) , .
$$
Any suggestions?
edit: You are right! I was a bit optimistic and also found a counterexample: Choosing $f_l(x)$ to become very close to $1$ very quickly and then only increasing slowly for a large interval makes the $f_l$ integral very large. On the other hand one can choose $f_s(x)$ negative and large enough, so that the integral for $f_s$ is very small. $f_s$ then increases just before $x_l$ to unity at $x_s=x_l$. This contribution should be small in comparison when the interval is large enough.
real-analysis functions definite-integrals
edited yesterday
Ethan Bolker
39k543102
asked 2 days ago
Diger
1,369411
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show 2 more comments
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0
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Let $f_s,f_l:(0,infty) mapsto {mathbb R}$ be monotonic increasing functions with $f_s(x) leq f_l(x)$ and let $x_s,x_l$ be the smallest positive roots of the equation $1-f_{s,l}(x)=0$ respectively (assuming they exist, are real and no double roots). Then I believe the following is true
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} geq int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}$$
and if it is true I would like to proof it.
I just don‘t know where to start.
Heuristically I think it is because the increase in integration interval $x_s geq x_l$ outweights the decrease of the integrand values $$frac{{rm d}x}{sqrt{1-f_s(x)}} leq frac{{rm d}x}{sqrt{1-f_l(x)}} , .$$
Observe for small deviations $f_l = f_s + delta f$ we have to first order for the RHS
$$
int_0^{x_l} frac{{rm d}x}{sqrt{1-f_s(x)}} left( 1 + frac{delta f(x)}{2(1-f_s(x))} right)
$$
and so this requires
$$
int_{x_l}^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} geq int_0^{x_l} frac{{rm d}x}{sqrt{1-f_s(x)}} left( frac{delta f(x)}{2(1-f_s(x))} right) , .
$$
Any suggestions?
edit: You are right! I was a bit optimistic and also found a counterexample: Choosing $f_l(x)$ to become very close to $1$ very quickly and then only increasing slowly for a large interval makes the $f_l$ integral very large. On the other hand one can choose $f_s(x)$ negative and large enough, so that the integral for $f_s$ is very small. $f_s$ then increases just before $x_l$ to unity at $x_s=x_l$. This contribution should be small in comparison when the interval is large enough.
real-analysis functions definite-integrals
edited yesterday
Ethan Bolker
39k543102
asked 2 days ago
Diger
1,369411
How is the rhs divergent for $f_l(x)=1-x^2$?
– Diger
2 days ago
Oops, you're right. Deleting. Sorry.
– Barry Cipra
2 days ago
Divergent integrals can not occur if $f$ is continuous and as mentioned no double roots occur.
– Diger
2 days ago
Ah, I think I meant $f_l(x)approx1-(1-x)^2$ for $xle1$. Then $int_0^1{dxoversqrt{1-f_l(x)}}approxint_0^1{dxover1-x}=infty$.
– Barry Cipra
2 days ago
By "$approx$" I mean replace the $2$ in $1-(1-x)^2$ with something slightly less than $2$. This eliminates the double root but keeps the integral as large as you want.
– Barry Cipra
2 days ago
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f_s,f_l:(0,infty) mapsto {mathbb R}$ be monotonic increasing functions with $f_s(x) leq f_l(x)$ and let $x_s,x_l$ be the smallest positive roots of the equation $1-f_{s,l}(x)=0$ respectively (assuming they exist, are real and no double roots). Then I believe the following is true
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} geq int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}$$
and if it is true I would like to proof it.
I just don‘t know where to start.
Heuristically I think it is because the increase in integration interval $x_s geq x_l$ outweights the decrease of the integrand values $$frac{{rm d}x}{sqrt{1-f_s(x)}} leq frac{{rm d}x}{sqrt{1-f_l(x)}} , .$$
Observe for small deviations $f_l = f_s + delta f$ we have to first order for the RHS
$$
int_0^{x_l} frac{{rm d}x}{sqrt{1-f_s(x)}} left( 1 + frac{delta f(x)}{2(1-f_s(x))} right)
$$
and so this requires
$$
int_{x_l}^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} geq int_0^{x_l} frac{{rm d}x}{sqrt{1-f_s(x)}} left( frac{delta f(x)}{2(1-f_s(x))} right) , .
$$
Any suggestions?
edit: You are right! I was a bit optimistic and also found a counterexample: Choosing $f_l(x)$ to become very close to $1$ very quickly and then only increasing slowly for a large interval makes the $f_l$ integral very large. On the other hand one can choose $f_s(x)$ negative and large enough, so that the integral for $f_s$ is very small. $f_s$ then increases just before $x_l$ to unity at $x_s=x_l$. This contribution should be small in comparison when the interval is large enough.
real-analysis functions definite-integrals
edited yesterday
Ethan Bolker
39k543102
asked 2 days ago
Diger
1,369411
Let $f_s,f_l:(0,infty) mapsto {mathbb R}$ be monotonic increasing functions with $f_s(x) leq f_l(x)$ and let $x_s,x_l$ be the smallest positive roots of the equation $1-f_{s,l}(x)=0$ respectively (assuming they exist, are real and no double roots). Then I believe the following is true
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} geq int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}$$
and if it is true I would like to proof it.
I just don‘t know where to start.
Heuristically I think it is because the increase in integration interval $x_s geq x_l$ outweights the decrease of the integrand values $$frac{{rm d}x}{sqrt{1-f_s(x)}} leq frac{{rm d}x}{sqrt{1-f_l(x)}} , .$$
Observe for small deviations $f_l = f_s + delta f$ we have to first order for the RHS
$$
int_0^{x_l} frac{{rm d}x}{sqrt{1-f_s(x)}} left( 1 + frac{delta f(x)}{2(1-f_s(x))} right)
$$
and so this requires
$$
int_{x_l}^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} geq int_0^{x_l} frac{{rm d}x}{sqrt{1-f_s(x)}} left( frac{delta f(x)}{2(1-f_s(x))} right) , .
$$
Any suggestions?
edit: You are right! I was a bit optimistic and also found a counterexample: Choosing $f_l(x)$ to become very close to $1$ very quickly and then only increasing slowly for a large interval makes the $f_l$ integral very large. On the other hand one can choose $f_s(x)$ negative and large enough, so that the integral for $f_s$ is very small. $f_s$ then increases just before $x_l$ to unity at $x_s=x_l$. This contribution should be small in comparison when the interval is large enough.
real-analysis functions definite-integrals
real-analysis functions definite-integrals
edited yesterday
Ethan Bolker
39k543102
asked 2 days ago
Diger
1,369411
edited yesterday
Ethan Bolker
39k543102
asked 2 days ago
Diger
1,369411
edited yesterday
Ethan Bolker
39k543102
edited yesterday
Ethan Bolker
39k543102
edited yesterday
Ethan Bolker
39k543102
39k543102
asked 2 days ago
Diger
1,369411
asked 2 days ago
Diger
1,369411
asked 2 days ago
Diger
1,369411
1,369411
How is the rhs divergent for $f_l(x)=1-x^2$?
– Diger
2 days ago
Oops, you're right. Deleting. Sorry.
– Barry Cipra
2 days ago
Divergent integrals can not occur if $f$ is continuous and as mentioned no double roots occur.
– Diger
2 days ago
Ah, I think I meant $f_l(x)approx1-(1-x)^2$ for $xle1$. Then $int_0^1{dxoversqrt{1-f_l(x)}}approxint_0^1{dxover1-x}=infty$.
– Barry Cipra
2 days ago
By "$approx$" I mean replace the $2$ in $1-(1-x)^2$ with something slightly less than $2$. This eliminates the double root but keeps the integral as large as you want.
– Barry Cipra
2 days ago
|
show 2 more comments
How is the rhs divergent for $f_l(x)=1-x^2$?
– Diger
2 days ago
Oops, you're right. Deleting. Sorry.
– Barry Cipra
2 days ago
Divergent integrals can not occur if $f$ is continuous and as mentioned no double roots occur.
– Diger
2 days ago
Ah, I think I meant $f_l(x)approx1-(1-x)^2$ for $xle1$. Then $int_0^1{dxoversqrt{1-f_l(x)}}approxint_0^1{dxover1-x}=infty$.
– Barry Cipra
2 days ago
By "$approx$" I mean replace the $2$ in $1-(1-x)^2$ with something slightly less than $2$. This eliminates the double root but keeps the integral as large as you want.
– Barry Cipra
2 days ago
How is the rhs divergent for $f_l(x)=1-x^2$?
– Diger
2 days ago
How is the rhs divergent for $f_l(x)=1-x^2$?
– Diger
2 days ago
Oops, you're right. Deleting. Sorry.
– Barry Cipra
2 days ago
Oops, you're right. Deleting. Sorry.
– Barry Cipra
2 days ago
Divergent integrals can not occur if $f$ is continuous and as mentioned no double roots occur.
– Diger
2 days ago
Divergent integrals can not occur if $f$ is continuous and as mentioned no double roots occur.
– Diger
2 days ago
Ah, I think I meant $f_l(x)approx1-(1-x)^2$ for $xle1$. Then $int_0^1{dxoversqrt{1-f_l(x)}}approxint_0^1{dxover1-x}=infty$.
– Barry Cipra
2 days ago
Ah, I think I meant $f_l(x)approx1-(1-x)^2$ for $xle1$. Then $int_0^1{dxoversqrt{1-f_l(x)}}approxint_0^1{dxover1-x}=infty$.
– Barry Cipra
2 days ago
By "$approx$" I mean replace the $2$ in $1-(1-x)^2$ with something slightly less than $2$. This eliminates the double root but keeps the integral as large as you want.
– Barry Cipra
2 days ago
By "$approx$" I mean replace the $2$ in $1-(1-x)^2$ with something slightly less than $2$. This eliminates the double root but keeps the integral as large as you want.
– Barry Cipra
2 days ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
Here is a counterexample with polynomials:
Let $f_l(x)={1over2}(x+x^2-x^3+x^4)$ and $f_s(x)=x^2$. We need to check that $f_l$ is increasing on $[0,infty)$ and that $f_l(x)ge f_s(x)$ for $xin[0,infty)$. Once we do that, it's clear that $x_l=x_s=1$, so that $int_0^{x_s}{dxoversqrt{1-f_s(x)}}ltint_0^{x_l}{dxoversqrt{1-f_l(x)}}$.
To show that $f_l(x)={1over2}(x+x^2-x^3+x^4)$ is increasing, we take the first two derivatives:
$$f_l'(x)={1over2}(1+2x-3x^2+4x^3)\
f_l''(x)={1over2}(2-6x+12x^2)=1-3x+6x^2$$
Since $3^2-4cdot1cdot6=-15lt0$, the second derivative is never $0$, hence, since $f_l''(0)=1$, it's always positive. This implies the first derivative is always increasing, so, since $f'(0)={1over2}$, the first derivative is positive for $xge0$. This in turn implies the function $f_l$ is increasing on $[0,infty)$.
To show that $f_l(x)ge f_s(x)$ for $xin[0,infty)$, we see that
$$f_l(x)-f_s(x)={1over2}(x-x^2-x^3+x^4)={1over2}x(1-x)^2(1+x)ge0$$
Just for fun, the integrals are
$$int_0^1{dxoversqrt{1-{1over2}(x+x^2-x^3+x^4)}}approx1.62868$$
(according to Wolfram Alpha) and
$$int_0^1{dxoversqrt{1-x^2}}={piover2}approx1.5708$$
Finally, if you want an example with $x_sgt x_l$, you can take $f_s(x)=(28x/29)^2$, in which case
$$int_0^{29/28}{dxoversqrt{1-left(28xover29right)^2}}={29piover56}approx1.6269$$
answered yesterday
Barry Cipra
58k652121
add a comment |
up vote
1
down vote
Define $f_s(x) = x-1,$ $f_l(x)=(x-1)/2$ for $xin [0,1],$ $f_l(x)=2(x-1)$ for $x>1.$ We then have $x_s=x_l =1.$ However
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} < int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}.$$
answered 2 days ago
zhw.
70.3k43075
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Here is a counterexample with polynomials:
Let $f_l(x)={1over2}(x+x^2-x^3+x^4)$ and $f_s(x)=x^2$. We need to check that $f_l$ is increasing on $[0,infty)$ and that $f_l(x)ge f_s(x)$ for $xin[0,infty)$. Once we do that, it's clear that $x_l=x_s=1$, so that $int_0^{x_s}{dxoversqrt{1-f_s(x)}}ltint_0^{x_l}{dxoversqrt{1-f_l(x)}}$.
To show that $f_l(x)={1over2}(x+x^2-x^3+x^4)$ is increasing, we take the first two derivatives:
$$f_l'(x)={1over2}(1+2x-3x^2+4x^3)\
f_l''(x)={1over2}(2-6x+12x^2)=1-3x+6x^2$$
Since $3^2-4cdot1cdot6=-15lt0$, the second derivative is never $0$, hence, since $f_l''(0)=1$, it's always positive. This implies the first derivative is always increasing, so, since $f'(0)={1over2}$, the first derivative is positive for $xge0$. This in turn implies the function $f_l$ is increasing on $[0,infty)$.
To show that $f_l(x)ge f_s(x)$ for $xin[0,infty)$, we see that
$$f_l(x)-f_s(x)={1over2}(x-x^2-x^3+x^4)={1over2}x(1-x)^2(1+x)ge0$$
Just for fun, the integrals are
$$int_0^1{dxoversqrt{1-{1over2}(x+x^2-x^3+x^4)}}approx1.62868$$
(according to Wolfram Alpha) and
$$int_0^1{dxoversqrt{1-x^2}}={piover2}approx1.5708$$
Finally, if you want an example with $x_sgt x_l$, you can take $f_s(x)=(28x/29)^2$, in which case
$$int_0^{29/28}{dxoversqrt{1-left(28xover29right)^2}}={29piover56}approx1.6269$$
answered yesterday
Barry Cipra
58k652121
add a comment |
up vote
2
down vote
Here is a counterexample with polynomials:
Let $f_l(x)={1over2}(x+x^2-x^3+x^4)$ and $f_s(x)=x^2$. We need to check that $f_l$ is increasing on $[0,infty)$ and that $f_l(x)ge f_s(x)$ for $xin[0,infty)$. Once we do that, it's clear that $x_l=x_s=1$, so that $int_0^{x_s}{dxoversqrt{1-f_s(x)}}ltint_0^{x_l}{dxoversqrt{1-f_l(x)}}$.
To show that $f_l(x)={1over2}(x+x^2-x^3+x^4)$ is increasing, we take the first two derivatives:
$$f_l'(x)={1over2}(1+2x-3x^2+4x^3)\
f_l''(x)={1over2}(2-6x+12x^2)=1-3x+6x^2$$
Since $3^2-4cdot1cdot6=-15lt0$, the second derivative is never $0$, hence, since $f_l''(0)=1$, it's always positive. This implies the first derivative is always increasing, so, since $f'(0)={1over2}$, the first derivative is positive for $xge0$. This in turn implies the function $f_l$ is increasing on $[0,infty)$.
To show that $f_l(x)ge f_s(x)$ for $xin[0,infty)$, we see that
$$f_l(x)-f_s(x)={1over2}(x-x^2-x^3+x^4)={1over2}x(1-x)^2(1+x)ge0$$
Just for fun, the integrals are
$$int_0^1{dxoversqrt{1-{1over2}(x+x^2-x^3+x^4)}}approx1.62868$$
(according to Wolfram Alpha) and
$$int_0^1{dxoversqrt{1-x^2}}={piover2}approx1.5708$$
Finally, if you want an example with $x_sgt x_l$, you can take $f_s(x)=(28x/29)^2$, in which case
$$int_0^{29/28}{dxoversqrt{1-left(28xover29right)^2}}={29piover56}approx1.6269$$
answered yesterday
Barry Cipra
58k652121
add a comment |
up vote
2
down vote
up vote
2
down vote
Here is a counterexample with polynomials:
Let $f_l(x)={1over2}(x+x^2-x^3+x^4)$ and $f_s(x)=x^2$. We need to check that $f_l$ is increasing on $[0,infty)$ and that $f_l(x)ge f_s(x)$ for $xin[0,infty)$. Once we do that, it's clear that $x_l=x_s=1$, so that $int_0^{x_s}{dxoversqrt{1-f_s(x)}}ltint_0^{x_l}{dxoversqrt{1-f_l(x)}}$.
To show that $f_l(x)={1over2}(x+x^2-x^3+x^4)$ is increasing, we take the first two derivatives:
$$f_l'(x)={1over2}(1+2x-3x^2+4x^3)\
f_l''(x)={1over2}(2-6x+12x^2)=1-3x+6x^2$$
Since $3^2-4cdot1cdot6=-15lt0$, the second derivative is never $0$, hence, since $f_l''(0)=1$, it's always positive. This implies the first derivative is always increasing, so, since $f'(0)={1over2}$, the first derivative is positive for $xge0$. This in turn implies the function $f_l$ is increasing on $[0,infty)$.
To show that $f_l(x)ge f_s(x)$ for $xin[0,infty)$, we see that
$$f_l(x)-f_s(x)={1over2}(x-x^2-x^3+x^4)={1over2}x(1-x)^2(1+x)ge0$$
Just for fun, the integrals are
$$int_0^1{dxoversqrt{1-{1over2}(x+x^2-x^3+x^4)}}approx1.62868$$
(according to Wolfram Alpha) and
$$int_0^1{dxoversqrt{1-x^2}}={piover2}approx1.5708$$
Finally, if you want an example with $x_sgt x_l$, you can take $f_s(x)=(28x/29)^2$, in which case
$$int_0^{29/28}{dxoversqrt{1-left(28xover29right)^2}}={29piover56}approx1.6269$$
answered yesterday
Barry Cipra
58k652121
Here is a counterexample with polynomials:
Let $f_l(x)={1over2}(x+x^2-x^3+x^4)$ and $f_s(x)=x^2$. We need to check that $f_l$ is increasing on $[0,infty)$ and that $f_l(x)ge f_s(x)$ for $xin[0,infty)$. Once we do that, it's clear that $x_l=x_s=1$, so that $int_0^{x_s}{dxoversqrt{1-f_s(x)}}ltint_0^{x_l}{dxoversqrt{1-f_l(x)}}$.
To show that $f_l(x)={1over2}(x+x^2-x^3+x^4)$ is increasing, we take the first two derivatives:
$$f_l'(x)={1over2}(1+2x-3x^2+4x^3)\
f_l''(x)={1over2}(2-6x+12x^2)=1-3x+6x^2$$
Since $3^2-4cdot1cdot6=-15lt0$, the second derivative is never $0$, hence, since $f_l''(0)=1$, it's always positive. This implies the first derivative is always increasing, so, since $f'(0)={1over2}$, the first derivative is positive for $xge0$. This in turn implies the function $f_l$ is increasing on $[0,infty)$.
To show that $f_l(x)ge f_s(x)$ for $xin[0,infty)$, we see that
$$f_l(x)-f_s(x)={1over2}(x-x^2-x^3+x^4)={1over2}x(1-x)^2(1+x)ge0$$
Just for fun, the integrals are
$$int_0^1{dxoversqrt{1-{1over2}(x+x^2-x^3+x^4)}}approx1.62868$$
(according to Wolfram Alpha) and
$$int_0^1{dxoversqrt{1-x^2}}={piover2}approx1.5708$$
Finally, if you want an example with $x_sgt x_l$, you can take $f_s(x)=(28x/29)^2$, in which case
$$int_0^{29/28}{dxoversqrt{1-left(28xover29right)^2}}={29piover56}approx1.6269$$
answered yesterday
Barry Cipra
58k652121
answered yesterday
Barry Cipra
58k652121
answered yesterday
Barry Cipra
58k652121
answered yesterday
Barry Cipra
58k652121
58k652121
add a comment |
add a comment |
up vote
1
down vote
Define $f_s(x) = x-1,$ $f_l(x)=(x-1)/2$ for $xin [0,1],$ $f_l(x)=2(x-1)$ for $x>1.$ We then have $x_s=x_l =1.$ However
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} < int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}.$$
answered 2 days ago
zhw.
70.3k43075
add a comment |
up vote
1
down vote
Define $f_s(x) = x-1,$ $f_l(x)=(x-1)/2$ for $xin [0,1],$ $f_l(x)=2(x-1)$ for $x>1.$ We then have $x_s=x_l =1.$ However
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} < int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}.$$
answered 2 days ago
zhw.
70.3k43075
add a comment |
up vote
1
down vote
up vote
1
down vote
Define $f_s(x) = x-1,$ $f_l(x)=(x-1)/2$ for $xin [0,1],$ $f_l(x)=2(x-1)$ for $x>1.$ We then have $x_s=x_l =1.$ However
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} < int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}.$$
answered 2 days ago
zhw.
70.3k43075
Define $f_s(x) = x-1,$ $f_l(x)=(x-1)/2$ for $xin [0,1],$ $f_l(x)=2(x-1)$ for $x>1.$ We then have $x_s=x_l =1.$ However
$$ int_0^{x_s} frac{{rm d}x}{sqrt{1-f_s(x)}} < int_0^{x_l} frac{{rm d}x}{sqrt{1-f_l(x)}}.$$
answered 2 days ago
zhw.
70.3k43075
answered 2 days ago
zhw.
70.3k43075
answered 2 days ago
zhw.
70.3k43075
answered 2 days ago
zhw.
70.3k43075
70.3k43075
add a comment |
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How is the rhs divergent for $f_l(x)=1-x^2$?
– Diger
2 days ago
Oops, you're right. Deleting. Sorry.
– Barry Cipra
2 days ago
Divergent integrals can not occur if $f$ is continuous and as mentioned no double roots occur.
– Diger
2 days ago
Ah, I think I meant $f_l(x)approx1-(1-x)^2$ for $xle1$. Then $int_0^1{dxoversqrt{1-f_l(x)}}approxint_0^1{dxover1-x}=infty$.
– Barry Cipra
2 days ago
By "$approx$" I mean replace the $2$ in $1-(1-x)^2$ with something slightly less than $2$. This eliminates the double root but keeps the integral as large as you want.
– Barry Cipra
2 days ago