Prove that $int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3$ for $m>3$.

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I've been trying to prove this inequality for a while.




$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3.$$
It can be rewritten in terms of the incomplete gamma function
$$Gamma(m+1,2(m-2))leqfrac(2(m-2))^m+1e^-2(m-2)m-3$$
or the generalized exponential integral
$$E_-m(2(m-2))leqfrace^-2(m-2)m-3.$$




For any version, integrating by parts gives a recurring expression, but it quickly becomes intractable.



The inequality can be directly checked for low values of $m$. It can also be proved that $(m-3)int_0^inftye^-2(m-2)s(1+s)^m ds$ converges to one as $m$ grows to infinity. But I have not been able to prove that the latter expression is always increasing in $m$.



Using the fact that $1+sleq e^s$ holds for every $sgeq 0$, we can see that the left hand side is smaller than $frac1m-4$, but I am interested in the tighter bound.







share|cite|improve this question






















  • Is $m$ actually an integer?
    – Diger
    Aug 19 at 0:14










  • You may assume that if that helps. Then the inequality can be written as $m!sum_k=0^mfrac(2(m-1))^kk!leqfrac(2(m-1))^m+1m-3$.
    – userq3125
    Aug 19 at 1:08











  • So you still need the proof?
    – Diger
    Aug 21 at 5:48










  • Yes, I do. Any ideas?
    – userq3125
    Aug 21 at 6:52














up vote
2
down vote

favorite
1












I've been trying to prove this inequality for a while.




$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3.$$
It can be rewritten in terms of the incomplete gamma function
$$Gamma(m+1,2(m-2))leqfrac(2(m-2))^m+1e^-2(m-2)m-3$$
or the generalized exponential integral
$$E_-m(2(m-2))leqfrace^-2(m-2)m-3.$$




For any version, integrating by parts gives a recurring expression, but it quickly becomes intractable.



The inequality can be directly checked for low values of $m$. It can also be proved that $(m-3)int_0^inftye^-2(m-2)s(1+s)^m ds$ converges to one as $m$ grows to infinity. But I have not been able to prove that the latter expression is always increasing in $m$.



Using the fact that $1+sleq e^s$ holds for every $sgeq 0$, we can see that the left hand side is smaller than $frac1m-4$, but I am interested in the tighter bound.







share|cite|improve this question






















  • Is $m$ actually an integer?
    – Diger
    Aug 19 at 0:14










  • You may assume that if that helps. Then the inequality can be written as $m!sum_k=0^mfrac(2(m-1))^kk!leqfrac(2(m-1))^m+1m-3$.
    – userq3125
    Aug 19 at 1:08











  • So you still need the proof?
    – Diger
    Aug 21 at 5:48










  • Yes, I do. Any ideas?
    – userq3125
    Aug 21 at 6:52












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I've been trying to prove this inequality for a while.




$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3.$$
It can be rewritten in terms of the incomplete gamma function
$$Gamma(m+1,2(m-2))leqfrac(2(m-2))^m+1e^-2(m-2)m-3$$
or the generalized exponential integral
$$E_-m(2(m-2))leqfrace^-2(m-2)m-3.$$




For any version, integrating by parts gives a recurring expression, but it quickly becomes intractable.



The inequality can be directly checked for low values of $m$. It can also be proved that $(m-3)int_0^inftye^-2(m-2)s(1+s)^m ds$ converges to one as $m$ grows to infinity. But I have not been able to prove that the latter expression is always increasing in $m$.



Using the fact that $1+sleq e^s$ holds for every $sgeq 0$, we can see that the left hand side is smaller than $frac1m-4$, but I am interested in the tighter bound.







share|cite|improve this question














I've been trying to prove this inequality for a while.




$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3.$$
It can be rewritten in terms of the incomplete gamma function
$$Gamma(m+1,2(m-2))leqfrac(2(m-2))^m+1e^-2(m-2)m-3$$
or the generalized exponential integral
$$E_-m(2(m-2))leqfrace^-2(m-2)m-3.$$




For any version, integrating by parts gives a recurring expression, but it quickly becomes intractable.



The inequality can be directly checked for low values of $m$. It can also be proved that $(m-3)int_0^inftye^-2(m-2)s(1+s)^m ds$ converges to one as $m$ grows to infinity. But I have not been able to prove that the latter expression is always increasing in $m$.



Using the fact that $1+sleq e^s$ holds for every $sgeq 0$, we can see that the left hand side is smaller than $frac1m-4$, but I am interested in the tighter bound.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 18 at 23:11

























asked Aug 18 at 1:33









userq3125

787




787











  • Is $m$ actually an integer?
    – Diger
    Aug 19 at 0:14










  • You may assume that if that helps. Then the inequality can be written as $m!sum_k=0^mfrac(2(m-1))^kk!leqfrac(2(m-1))^m+1m-3$.
    – userq3125
    Aug 19 at 1:08











  • So you still need the proof?
    – Diger
    Aug 21 at 5:48










  • Yes, I do. Any ideas?
    – userq3125
    Aug 21 at 6:52
















  • Is $m$ actually an integer?
    – Diger
    Aug 19 at 0:14










  • You may assume that if that helps. Then the inequality can be written as $m!sum_k=0^mfrac(2(m-1))^kk!leqfrac(2(m-1))^m+1m-3$.
    – userq3125
    Aug 19 at 1:08











  • So you still need the proof?
    – Diger
    Aug 21 at 5:48










  • Yes, I do. Any ideas?
    – userq3125
    Aug 21 at 6:52















Is $m$ actually an integer?
– Diger
Aug 19 at 0:14




Is $m$ actually an integer?
– Diger
Aug 19 at 0:14












You may assume that if that helps. Then the inequality can be written as $m!sum_k=0^mfrac(2(m-1))^kk!leqfrac(2(m-1))^m+1m-3$.
– userq3125
Aug 19 at 1:08





You may assume that if that helps. Then the inequality can be written as $m!sum_k=0^mfrac(2(m-1))^kk!leqfrac(2(m-1))^m+1m-3$.
– userq3125
Aug 19 at 1:08













So you still need the proof?
– Diger
Aug 21 at 5:48




So you still need the proof?
– Diger
Aug 21 at 5:48












Yes, I do. Any ideas?
– userq3125
Aug 21 at 6:52




Yes, I do. Any ideas?
– userq3125
Aug 21 at 6:52










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










On $(3,infty)$ we have
beginalign
0 leq &(m-3)int_0^infty e^-2(m-2)s (1+s)^m , rm ds \
stackrels=e^t-1= &(m-3)int_0^infty e^-2(m-2)left(e^t-1right)+(m+1)t , rm dt \
leq &(m-3) int_0^infty e^-2(m-2)left(t+fract^22+fract^36right)+(m+1)t , rm dt \
stackrelu=(t+1)^3= &(m-3) , underbracee^-3int_1^infty frace^-fracm-23(u-1)+3u^1/33u^2/3 , rm du_equiv I_m , .
endalign
Now
beginalign
lim_mrightarrow infty (m-3)I_m &= lim_mrightarrow infty fracm-3m-2 , e^-3 sum_k=0^infty frac3^kk! left(fracm-23right)^-frack+13+1 e^fracm-23 , Gammaleft(frack+13 , fracm-23right) \
&=1
endalign
as can be seen by expanding $e^3u^1/3$.



It therefore remains to show that $(m-3)I_m$ is an increasing function. For that the idea is to remove the $m$-dependent pre-factor before taking the derivative to keep it tractable. In fact we have the identity
$$
(m-2)I_m = 1 + e^-3 int_1^infty left u^-4/3 - frac2 u^-5/33 right , e^-fracm-23(u-1)+3u^1/3 , rm du
$$
by partial integration. Hence
beginalign
fracrm drm dm (m-3)I_m &= fracrm drm dm (m-2)I_m - fracrm drm dm I_m \
&=e^-3 int_1^infty underbraceleft fracu^1/39 - fracu^-1/33 + fracu^-2/39 + fracu^-4/33 - frac2 u^-5/39 right_equiv f(u) , e^-fracm-23(u-1)+3u^1/3 , rm du \
&geq 0
endalign
because $f(u)geq 0$ which can be seen as follows:



First $f(1)=0$ and
$$f'(u) = fracu^-2/327 + fracu^-4/39 - frac2u^-5/327 - frac4u^-7/39 + frac10u^-8/327 geq 0 , .$$



The latter because $f'(1)=0$ and $f'(infty)=0$ while
$$ f''(u)=-frac2u^-5/381 - frac4u^-7/327 + frac10u^-8/381 + frac28u^-10/327 - frac80u^-11/381 = 0 $$
has only $1$ real-valued solution $u_0=z^3>1$ where $z$ is the only real solution of
$$
z^5 + z^4 + 7z^3 + 2z^2 + 2z - 40 = 0
$$
which corresponds to a maximum of $f'(u)$, since
beginalign
f'''(u_0)&=frac10u_0^-8/3243 + frac28u_0^-10/381 - frac80u_0^-11/3243 - frac280u_0^-13/381 + frac880u_0^-14/3243 \
&approx -0.002446492623 < 0
endalign
with $u_0 approx 3.101517308$.






share|cite|improve this answer






















  • Really like the flow of the proof. Could you add details on how you found $lim_m->infty(m-3)I_m=1$? Thanks.
    – skbmoore
    Aug 21 at 19:34










  • Do you mean how I arrived at the series expansion, or why the limit using the series is $1$?
    – Diger
    Aug 21 at 20:04










  • Thanks a lot, that was extremely helpful! Perhaps the limiting result can be more easily seen using the expression that you found for $(m-2)I_m$.
    – userq3125
    Aug 22 at 8:00










  • Ah indeed ;) That's much simpler.
    – Diger
    Aug 22 at 8:09






  • 1




    It also seems to be the best bound, since any value smaller $3$ violates the inequality for large enough $m$. In fact it seems that $$(m-2) int_0^inftye^-2(m-2)s(1+s)^m , rm ds geq 1$$ for any $m>2$.
    – Diger
    Aug 22 at 9:08

















up vote
1
down vote













$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3$$



put $1+s=t$ thus $ds=dt$



also as $s$ changes from 0 to $infty$ thus $t$ changes from 1 to $infty$.



$$I=int_1^inftye^-2(m-2)(t-1)t^m dt=e^2(m-2)int_1^inftye^-2(m-2)tt^m dt$$



put $2(m-2)t=x$ thus $2(m-2)dt=dx$



$x$ changes from $2(m-2)$ to $infty$



$$I=frace^2(m-2)2^m+1(m-2)^m+1int_2(m-2)^infty e^-xx^m dx$$
$$Gamma(m+1)=int_0^infty e^-t t^m dt$$






share|cite|improve this answer






















  • As $s$ changes from $0$ to infinity, $t$ changes from $1$ to infinity.
    – Ahmad Bazzi
    Aug 18 at 1:46











  • yes it was a mistake I corrected it
    – James
    Aug 18 at 1:48










  • I added the version of the inequality for the incomplete gamma function that can be obtained using your derivation.
    – userq3125
    Aug 18 at 5:46










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










On $(3,infty)$ we have
beginalign
0 leq &(m-3)int_0^infty e^-2(m-2)s (1+s)^m , rm ds \
stackrels=e^t-1= &(m-3)int_0^infty e^-2(m-2)left(e^t-1right)+(m+1)t , rm dt \
leq &(m-3) int_0^infty e^-2(m-2)left(t+fract^22+fract^36right)+(m+1)t , rm dt \
stackrelu=(t+1)^3= &(m-3) , underbracee^-3int_1^infty frace^-fracm-23(u-1)+3u^1/33u^2/3 , rm du_equiv I_m , .
endalign
Now
beginalign
lim_mrightarrow infty (m-3)I_m &= lim_mrightarrow infty fracm-3m-2 , e^-3 sum_k=0^infty frac3^kk! left(fracm-23right)^-frack+13+1 e^fracm-23 , Gammaleft(frack+13 , fracm-23right) \
&=1
endalign
as can be seen by expanding $e^3u^1/3$.



It therefore remains to show that $(m-3)I_m$ is an increasing function. For that the idea is to remove the $m$-dependent pre-factor before taking the derivative to keep it tractable. In fact we have the identity
$$
(m-2)I_m = 1 + e^-3 int_1^infty left u^-4/3 - frac2 u^-5/33 right , e^-fracm-23(u-1)+3u^1/3 , rm du
$$
by partial integration. Hence
beginalign
fracrm drm dm (m-3)I_m &= fracrm drm dm (m-2)I_m - fracrm drm dm I_m \
&=e^-3 int_1^infty underbraceleft fracu^1/39 - fracu^-1/33 + fracu^-2/39 + fracu^-4/33 - frac2 u^-5/39 right_equiv f(u) , e^-fracm-23(u-1)+3u^1/3 , rm du \
&geq 0
endalign
because $f(u)geq 0$ which can be seen as follows:



First $f(1)=0$ and
$$f'(u) = fracu^-2/327 + fracu^-4/39 - frac2u^-5/327 - frac4u^-7/39 + frac10u^-8/327 geq 0 , .$$



The latter because $f'(1)=0$ and $f'(infty)=0$ while
$$ f''(u)=-frac2u^-5/381 - frac4u^-7/327 + frac10u^-8/381 + frac28u^-10/327 - frac80u^-11/381 = 0 $$
has only $1$ real-valued solution $u_0=z^3>1$ where $z$ is the only real solution of
$$
z^5 + z^4 + 7z^3 + 2z^2 + 2z - 40 = 0
$$
which corresponds to a maximum of $f'(u)$, since
beginalign
f'''(u_0)&=frac10u_0^-8/3243 + frac28u_0^-10/381 - frac80u_0^-11/3243 - frac280u_0^-13/381 + frac880u_0^-14/3243 \
&approx -0.002446492623 < 0
endalign
with $u_0 approx 3.101517308$.






share|cite|improve this answer






















  • Really like the flow of the proof. Could you add details on how you found $lim_m->infty(m-3)I_m=1$? Thanks.
    – skbmoore
    Aug 21 at 19:34










  • Do you mean how I arrived at the series expansion, or why the limit using the series is $1$?
    – Diger
    Aug 21 at 20:04










  • Thanks a lot, that was extremely helpful! Perhaps the limiting result can be more easily seen using the expression that you found for $(m-2)I_m$.
    – userq3125
    Aug 22 at 8:00










  • Ah indeed ;) That's much simpler.
    – Diger
    Aug 22 at 8:09






  • 1




    It also seems to be the best bound, since any value smaller $3$ violates the inequality for large enough $m$. In fact it seems that $$(m-2) int_0^inftye^-2(m-2)s(1+s)^m , rm ds geq 1$$ for any $m>2$.
    – Diger
    Aug 22 at 9:08














up vote
2
down vote



accepted










On $(3,infty)$ we have
beginalign
0 leq &(m-3)int_0^infty e^-2(m-2)s (1+s)^m , rm ds \
stackrels=e^t-1= &(m-3)int_0^infty e^-2(m-2)left(e^t-1right)+(m+1)t , rm dt \
leq &(m-3) int_0^infty e^-2(m-2)left(t+fract^22+fract^36right)+(m+1)t , rm dt \
stackrelu=(t+1)^3= &(m-3) , underbracee^-3int_1^infty frace^-fracm-23(u-1)+3u^1/33u^2/3 , rm du_equiv I_m , .
endalign
Now
beginalign
lim_mrightarrow infty (m-3)I_m &= lim_mrightarrow infty fracm-3m-2 , e^-3 sum_k=0^infty frac3^kk! left(fracm-23right)^-frack+13+1 e^fracm-23 , Gammaleft(frack+13 , fracm-23right) \
&=1
endalign
as can be seen by expanding $e^3u^1/3$.



It therefore remains to show that $(m-3)I_m$ is an increasing function. For that the idea is to remove the $m$-dependent pre-factor before taking the derivative to keep it tractable. In fact we have the identity
$$
(m-2)I_m = 1 + e^-3 int_1^infty left u^-4/3 - frac2 u^-5/33 right , e^-fracm-23(u-1)+3u^1/3 , rm du
$$
by partial integration. Hence
beginalign
fracrm drm dm (m-3)I_m &= fracrm drm dm (m-2)I_m - fracrm drm dm I_m \
&=e^-3 int_1^infty underbraceleft fracu^1/39 - fracu^-1/33 + fracu^-2/39 + fracu^-4/33 - frac2 u^-5/39 right_equiv f(u) , e^-fracm-23(u-1)+3u^1/3 , rm du \
&geq 0
endalign
because $f(u)geq 0$ which can be seen as follows:



First $f(1)=0$ and
$$f'(u) = fracu^-2/327 + fracu^-4/39 - frac2u^-5/327 - frac4u^-7/39 + frac10u^-8/327 geq 0 , .$$



The latter because $f'(1)=0$ and $f'(infty)=0$ while
$$ f''(u)=-frac2u^-5/381 - frac4u^-7/327 + frac10u^-8/381 + frac28u^-10/327 - frac80u^-11/381 = 0 $$
has only $1$ real-valued solution $u_0=z^3>1$ where $z$ is the only real solution of
$$
z^5 + z^4 + 7z^3 + 2z^2 + 2z - 40 = 0
$$
which corresponds to a maximum of $f'(u)$, since
beginalign
f'''(u_0)&=frac10u_0^-8/3243 + frac28u_0^-10/381 - frac80u_0^-11/3243 - frac280u_0^-13/381 + frac880u_0^-14/3243 \
&approx -0.002446492623 < 0
endalign
with $u_0 approx 3.101517308$.






share|cite|improve this answer






















  • Really like the flow of the proof. Could you add details on how you found $lim_m->infty(m-3)I_m=1$? Thanks.
    – skbmoore
    Aug 21 at 19:34










  • Do you mean how I arrived at the series expansion, or why the limit using the series is $1$?
    – Diger
    Aug 21 at 20:04










  • Thanks a lot, that was extremely helpful! Perhaps the limiting result can be more easily seen using the expression that you found for $(m-2)I_m$.
    – userq3125
    Aug 22 at 8:00










  • Ah indeed ;) That's much simpler.
    – Diger
    Aug 22 at 8:09






  • 1




    It also seems to be the best bound, since any value smaller $3$ violates the inequality for large enough $m$. In fact it seems that $$(m-2) int_0^inftye^-2(m-2)s(1+s)^m , rm ds geq 1$$ for any $m>2$.
    – Diger
    Aug 22 at 9:08












up vote
2
down vote



accepted







up vote
2
down vote



accepted






On $(3,infty)$ we have
beginalign
0 leq &(m-3)int_0^infty e^-2(m-2)s (1+s)^m , rm ds \
stackrels=e^t-1= &(m-3)int_0^infty e^-2(m-2)left(e^t-1right)+(m+1)t , rm dt \
leq &(m-3) int_0^infty e^-2(m-2)left(t+fract^22+fract^36right)+(m+1)t , rm dt \
stackrelu=(t+1)^3= &(m-3) , underbracee^-3int_1^infty frace^-fracm-23(u-1)+3u^1/33u^2/3 , rm du_equiv I_m , .
endalign
Now
beginalign
lim_mrightarrow infty (m-3)I_m &= lim_mrightarrow infty fracm-3m-2 , e^-3 sum_k=0^infty frac3^kk! left(fracm-23right)^-frack+13+1 e^fracm-23 , Gammaleft(frack+13 , fracm-23right) \
&=1
endalign
as can be seen by expanding $e^3u^1/3$.



It therefore remains to show that $(m-3)I_m$ is an increasing function. For that the idea is to remove the $m$-dependent pre-factor before taking the derivative to keep it tractable. In fact we have the identity
$$
(m-2)I_m = 1 + e^-3 int_1^infty left u^-4/3 - frac2 u^-5/33 right , e^-fracm-23(u-1)+3u^1/3 , rm du
$$
by partial integration. Hence
beginalign
fracrm drm dm (m-3)I_m &= fracrm drm dm (m-2)I_m - fracrm drm dm I_m \
&=e^-3 int_1^infty underbraceleft fracu^1/39 - fracu^-1/33 + fracu^-2/39 + fracu^-4/33 - frac2 u^-5/39 right_equiv f(u) , e^-fracm-23(u-1)+3u^1/3 , rm du \
&geq 0
endalign
because $f(u)geq 0$ which can be seen as follows:



First $f(1)=0$ and
$$f'(u) = fracu^-2/327 + fracu^-4/39 - frac2u^-5/327 - frac4u^-7/39 + frac10u^-8/327 geq 0 , .$$



The latter because $f'(1)=0$ and $f'(infty)=0$ while
$$ f''(u)=-frac2u^-5/381 - frac4u^-7/327 + frac10u^-8/381 + frac28u^-10/327 - frac80u^-11/381 = 0 $$
has only $1$ real-valued solution $u_0=z^3>1$ where $z$ is the only real solution of
$$
z^5 + z^4 + 7z^3 + 2z^2 + 2z - 40 = 0
$$
which corresponds to a maximum of $f'(u)$, since
beginalign
f'''(u_0)&=frac10u_0^-8/3243 + frac28u_0^-10/381 - frac80u_0^-11/3243 - frac280u_0^-13/381 + frac880u_0^-14/3243 \
&approx -0.002446492623 < 0
endalign
with $u_0 approx 3.101517308$.






share|cite|improve this answer














On $(3,infty)$ we have
beginalign
0 leq &(m-3)int_0^infty e^-2(m-2)s (1+s)^m , rm ds \
stackrels=e^t-1= &(m-3)int_0^infty e^-2(m-2)left(e^t-1right)+(m+1)t , rm dt \
leq &(m-3) int_0^infty e^-2(m-2)left(t+fract^22+fract^36right)+(m+1)t , rm dt \
stackrelu=(t+1)^3= &(m-3) , underbracee^-3int_1^infty frace^-fracm-23(u-1)+3u^1/33u^2/3 , rm du_equiv I_m , .
endalign
Now
beginalign
lim_mrightarrow infty (m-3)I_m &= lim_mrightarrow infty fracm-3m-2 , e^-3 sum_k=0^infty frac3^kk! left(fracm-23right)^-frack+13+1 e^fracm-23 , Gammaleft(frack+13 , fracm-23right) \
&=1
endalign
as can be seen by expanding $e^3u^1/3$.



It therefore remains to show that $(m-3)I_m$ is an increasing function. For that the idea is to remove the $m$-dependent pre-factor before taking the derivative to keep it tractable. In fact we have the identity
$$
(m-2)I_m = 1 + e^-3 int_1^infty left u^-4/3 - frac2 u^-5/33 right , e^-fracm-23(u-1)+3u^1/3 , rm du
$$
by partial integration. Hence
beginalign
fracrm drm dm (m-3)I_m &= fracrm drm dm (m-2)I_m - fracrm drm dm I_m \
&=e^-3 int_1^infty underbraceleft fracu^1/39 - fracu^-1/33 + fracu^-2/39 + fracu^-4/33 - frac2 u^-5/39 right_equiv f(u) , e^-fracm-23(u-1)+3u^1/3 , rm du \
&geq 0
endalign
because $f(u)geq 0$ which can be seen as follows:



First $f(1)=0$ and
$$f'(u) = fracu^-2/327 + fracu^-4/39 - frac2u^-5/327 - frac4u^-7/39 + frac10u^-8/327 geq 0 , .$$



The latter because $f'(1)=0$ and $f'(infty)=0$ while
$$ f''(u)=-frac2u^-5/381 - frac4u^-7/327 + frac10u^-8/381 + frac28u^-10/327 - frac80u^-11/381 = 0 $$
has only $1$ real-valued solution $u_0=z^3>1$ where $z$ is the only real solution of
$$
z^5 + z^4 + 7z^3 + 2z^2 + 2z - 40 = 0
$$
which corresponds to a maximum of $f'(u)$, since
beginalign
f'''(u_0)&=frac10u_0^-8/3243 + frac28u_0^-10/381 - frac80u_0^-11/3243 - frac280u_0^-13/381 + frac880u_0^-14/3243 \
&approx -0.002446492623 < 0
endalign
with $u_0 approx 3.101517308$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 21 at 14:46

























answered Aug 21 at 14:02









Diger

66329




66329











  • Really like the flow of the proof. Could you add details on how you found $lim_m->infty(m-3)I_m=1$? Thanks.
    – skbmoore
    Aug 21 at 19:34










  • Do you mean how I arrived at the series expansion, or why the limit using the series is $1$?
    – Diger
    Aug 21 at 20:04










  • Thanks a lot, that was extremely helpful! Perhaps the limiting result can be more easily seen using the expression that you found for $(m-2)I_m$.
    – userq3125
    Aug 22 at 8:00










  • Ah indeed ;) That's much simpler.
    – Diger
    Aug 22 at 8:09






  • 1




    It also seems to be the best bound, since any value smaller $3$ violates the inequality for large enough $m$. In fact it seems that $$(m-2) int_0^inftye^-2(m-2)s(1+s)^m , rm ds geq 1$$ for any $m>2$.
    – Diger
    Aug 22 at 9:08
















  • Really like the flow of the proof. Could you add details on how you found $lim_m->infty(m-3)I_m=1$? Thanks.
    – skbmoore
    Aug 21 at 19:34










  • Do you mean how I arrived at the series expansion, or why the limit using the series is $1$?
    – Diger
    Aug 21 at 20:04










  • Thanks a lot, that was extremely helpful! Perhaps the limiting result can be more easily seen using the expression that you found for $(m-2)I_m$.
    – userq3125
    Aug 22 at 8:00










  • Ah indeed ;) That's much simpler.
    – Diger
    Aug 22 at 8:09






  • 1




    It also seems to be the best bound, since any value smaller $3$ violates the inequality for large enough $m$. In fact it seems that $$(m-2) int_0^inftye^-2(m-2)s(1+s)^m , rm ds geq 1$$ for any $m>2$.
    – Diger
    Aug 22 at 9:08















Really like the flow of the proof. Could you add details on how you found $lim_m->infty(m-3)I_m=1$? Thanks.
– skbmoore
Aug 21 at 19:34




Really like the flow of the proof. Could you add details on how you found $lim_m->infty(m-3)I_m=1$? Thanks.
– skbmoore
Aug 21 at 19:34












Do you mean how I arrived at the series expansion, or why the limit using the series is $1$?
– Diger
Aug 21 at 20:04




Do you mean how I arrived at the series expansion, or why the limit using the series is $1$?
– Diger
Aug 21 at 20:04












Thanks a lot, that was extremely helpful! Perhaps the limiting result can be more easily seen using the expression that you found for $(m-2)I_m$.
– userq3125
Aug 22 at 8:00




Thanks a lot, that was extremely helpful! Perhaps the limiting result can be more easily seen using the expression that you found for $(m-2)I_m$.
– userq3125
Aug 22 at 8:00












Ah indeed ;) That's much simpler.
– Diger
Aug 22 at 8:09




Ah indeed ;) That's much simpler.
– Diger
Aug 22 at 8:09




1




1




It also seems to be the best bound, since any value smaller $3$ violates the inequality for large enough $m$. In fact it seems that $$(m-2) int_0^inftye^-2(m-2)s(1+s)^m , rm ds geq 1$$ for any $m>2$.
– Diger
Aug 22 at 9:08




It also seems to be the best bound, since any value smaller $3$ violates the inequality for large enough $m$. In fact it seems that $$(m-2) int_0^inftye^-2(m-2)s(1+s)^m , rm ds geq 1$$ for any $m>2$.
– Diger
Aug 22 at 9:08










up vote
1
down vote













$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3$$



put $1+s=t$ thus $ds=dt$



also as $s$ changes from 0 to $infty$ thus $t$ changes from 1 to $infty$.



$$I=int_1^inftye^-2(m-2)(t-1)t^m dt=e^2(m-2)int_1^inftye^-2(m-2)tt^m dt$$



put $2(m-2)t=x$ thus $2(m-2)dt=dx$



$x$ changes from $2(m-2)$ to $infty$



$$I=frace^2(m-2)2^m+1(m-2)^m+1int_2(m-2)^infty e^-xx^m dx$$
$$Gamma(m+1)=int_0^infty e^-t t^m dt$$






share|cite|improve this answer






















  • As $s$ changes from $0$ to infinity, $t$ changes from $1$ to infinity.
    – Ahmad Bazzi
    Aug 18 at 1:46











  • yes it was a mistake I corrected it
    – James
    Aug 18 at 1:48










  • I added the version of the inequality for the incomplete gamma function that can be obtained using your derivation.
    – userq3125
    Aug 18 at 5:46














up vote
1
down vote













$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3$$



put $1+s=t$ thus $ds=dt$



also as $s$ changes from 0 to $infty$ thus $t$ changes from 1 to $infty$.



$$I=int_1^inftye^-2(m-2)(t-1)t^m dt=e^2(m-2)int_1^inftye^-2(m-2)tt^m dt$$



put $2(m-2)t=x$ thus $2(m-2)dt=dx$



$x$ changes from $2(m-2)$ to $infty$



$$I=frace^2(m-2)2^m+1(m-2)^m+1int_2(m-2)^infty e^-xx^m dx$$
$$Gamma(m+1)=int_0^infty e^-t t^m dt$$






share|cite|improve this answer






















  • As $s$ changes from $0$ to infinity, $t$ changes from $1$ to infinity.
    – Ahmad Bazzi
    Aug 18 at 1:46











  • yes it was a mistake I corrected it
    – James
    Aug 18 at 1:48










  • I added the version of the inequality for the incomplete gamma function that can be obtained using your derivation.
    – userq3125
    Aug 18 at 5:46












up vote
1
down vote










up vote
1
down vote









$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3$$



put $1+s=t$ thus $ds=dt$



also as $s$ changes from 0 to $infty$ thus $t$ changes from 1 to $infty$.



$$I=int_1^inftye^-2(m-2)(t-1)t^m dt=e^2(m-2)int_1^inftye^-2(m-2)tt^m dt$$



put $2(m-2)t=x$ thus $2(m-2)dt=dx$



$x$ changes from $2(m-2)$ to $infty$



$$I=frace^2(m-2)2^m+1(m-2)^m+1int_2(m-2)^infty e^-xx^m dx$$
$$Gamma(m+1)=int_0^infty e^-t t^m dt$$






share|cite|improve this answer














$$int_0^inftye^-2(m-2)s(1+s)^m dsleqfrac1m-3text for m>3$$



put $1+s=t$ thus $ds=dt$



also as $s$ changes from 0 to $infty$ thus $t$ changes from 1 to $infty$.



$$I=int_1^inftye^-2(m-2)(t-1)t^m dt=e^2(m-2)int_1^inftye^-2(m-2)tt^m dt$$



put $2(m-2)t=x$ thus $2(m-2)dt=dx$



$x$ changes from $2(m-2)$ to $infty$



$$I=frace^2(m-2)2^m+1(m-2)^m+1int_2(m-2)^infty e^-xx^m dx$$
$$Gamma(m+1)=int_0^infty e^-t t^m dt$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 18 at 1:59

























answered Aug 18 at 1:44









James

2,188619




2,188619











  • As $s$ changes from $0$ to infinity, $t$ changes from $1$ to infinity.
    – Ahmad Bazzi
    Aug 18 at 1:46











  • yes it was a mistake I corrected it
    – James
    Aug 18 at 1:48










  • I added the version of the inequality for the incomplete gamma function that can be obtained using your derivation.
    – userq3125
    Aug 18 at 5:46
















  • As $s$ changes from $0$ to infinity, $t$ changes from $1$ to infinity.
    – Ahmad Bazzi
    Aug 18 at 1:46











  • yes it was a mistake I corrected it
    – James
    Aug 18 at 1:48










  • I added the version of the inequality for the incomplete gamma function that can be obtained using your derivation.
    – userq3125
    Aug 18 at 5:46















As $s$ changes from $0$ to infinity, $t$ changes from $1$ to infinity.
– Ahmad Bazzi
Aug 18 at 1:46





As $s$ changes from $0$ to infinity, $t$ changes from $1$ to infinity.
– Ahmad Bazzi
Aug 18 at 1:46













yes it was a mistake I corrected it
– James
Aug 18 at 1:48




yes it was a mistake I corrected it
– James
Aug 18 at 1:48












I added the version of the inequality for the incomplete gamma function that can be obtained using your derivation.
– userq3125
Aug 18 at 5:46




I added the version of the inequality for the incomplete gamma function that can be obtained using your derivation.
– userq3125
Aug 18 at 5:46












 

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