Bound on log integral The 2019 Stack Overflow Developer Survey Results Are In ...

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Bound on log integral



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For what values of $pinmathbb{R}$ does the series $sum^{infty}_{n=4}{frac{1}{nlog (n)log( log(n))^p}}$ converge?Positive $T$ satisfying $int_T^infty x^{-log{x}} dx = int_0^Tx^{-log{x}}dx$Best upper bound on the number of divisors of $n$ that are larger than $N$.Limit of $F( Log( G(x) ) )$Integral $int_t^Tfrac{1}{phi-psi e^{-gamma(T-s)}} operatorname d !s$An upper bound of $ left| frac{1}{s}logzeta(s) right| $ for $Re s=sigma>1$, from this integral formula and a related comparisonEvaluate $lim_{ntoinfty}sum_{k=1}^n frac{log(k)}{nk}$On the integral $int_{e}^{infty}frac{t^{1/2}}{log^{1/2}left(tright)}alpha^{-t/logleft(tright)}dt,,alpha>1.$Continuity of $xlog(x^2)$ when $xneq 0$Asymptotics for $log$ integral












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$begingroup$


I am looking for an explanation of the bound
$$frac{1}{2pi}left(-frac{T log T}{1+(t-T)^2} - 2 int_T^infty frac{x log x (t-x)}{(1+(t-x)^2)^2} dx right)ll left( frac{1}{t+1} + frac{1}{T-t+1}right) log T,$$



where $t in [0,T].$



This is not for homework. Instead, this is a question from my senior thesis reading, so full answers are appreciated. I can bound the first term by $frac{log T}{T - t +1}$, but I am having trouble bounding the second.










share|cite|improve this question











$endgroup$












  • $begingroup$
    just bumping this.
    $endgroup$
    – William Chang
    Mar 23 at 2:05
















0












$begingroup$


I am looking for an explanation of the bound
$$frac{1}{2pi}left(-frac{T log T}{1+(t-T)^2} - 2 int_T^infty frac{x log x (t-x)}{(1+(t-x)^2)^2} dx right)ll left( frac{1}{t+1} + frac{1}{T-t+1}right) log T,$$



where $t in [0,T].$



This is not for homework. Instead, this is a question from my senior thesis reading, so full answers are appreciated. I can bound the first term by $frac{log T}{T - t +1}$, but I am having trouble bounding the second.










share|cite|improve this question











$endgroup$












  • $begingroup$
    just bumping this.
    $endgroup$
    – William Chang
    Mar 23 at 2:05














0












0








0





$begingroup$


I am looking for an explanation of the bound
$$frac{1}{2pi}left(-frac{T log T}{1+(t-T)^2} - 2 int_T^infty frac{x log x (t-x)}{(1+(t-x)^2)^2} dx right)ll left( frac{1}{t+1} + frac{1}{T-t+1}right) log T,$$



where $t in [0,T].$



This is not for homework. Instead, this is a question from my senior thesis reading, so full answers are appreciated. I can bound the first term by $frac{log T}{T - t +1}$, but I am having trouble bounding the second.










share|cite|improve this question











$endgroup$




I am looking for an explanation of the bound
$$frac{1}{2pi}left(-frac{T log T}{1+(t-T)^2} - 2 int_T^infty frac{x log x (t-x)}{(1+(t-x)^2)^2} dx right)ll left( frac{1}{t+1} + frac{1}{T-t+1}right) log T,$$



where $t in [0,T].$



This is not for homework. Instead, this is a question from my senior thesis reading, so full answers are appreciated. I can bound the first term by $frac{log T}{T - t +1}$, but I am having trouble bounding the second.







real-analysis integration analytic-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 2:06







William Chang

















asked Mar 22 at 12:35









William ChangWilliam Chang

1,81111227




1,81111227












  • $begingroup$
    just bumping this.
    $endgroup$
    – William Chang
    Mar 23 at 2:05


















  • $begingroup$
    just bumping this.
    $endgroup$
    – William Chang
    Mar 23 at 2:05
















$begingroup$
just bumping this.
$endgroup$
– William Chang
Mar 23 at 2:05




$begingroup$
just bumping this.
$endgroup$
– William Chang
Mar 23 at 2:05










0






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