Estimating $left |log(z) right |$ to calculate an integral?
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I'm doing the classical example of treating the integral $int_{0}^{infty }frac{log(x)}{left (1+x^2 right )^2}dx$ using residue theorem on the function $$f(z)=frac{log(z)}{left (1+z^2 right )^2}$$
I consider the contour $gamma$ comprising two semi-circular arcs with radii $R$ and $epsilon$ (having redefined the logarithm by deleting the negative imaginary axis).
I'm having trouble with estimating the part of $gamma$ on the semi-circle $left | z right |=R$. I have:
$$left |int_{zin gamma ,left | z right |=R}f(z) right |leq int_{0}^{pi}left | f(Re^{itheta })iRe^{itheta } right |dthetaleq int_{0}^{pi}R frac{sqrt{log^{2}(R)+theta ^{2}}}{(R^{2}-1)^{2}}dtheta $$
My book does the estimation:
$$left | f(z) right |leq frac{2log(R)}{(R^{2}-1)^{2}}$$
How did they get the $2log(R)$ estimation?
Thank's in advance.
complex-analysis contour-integration
edited Dec 30 '18 at 14:14
mrtaurho
6,12251641
asked Dec 30 '18 at 13:55
John11John11
1,0421821
$endgroup$
add a comment |
$begingroup$
I'm doing the classical example of treating the integral $int_{0}^{infty }frac{log(x)}{left (1+x^2 right )^2}dx$ using residue theorem on the function $$f(z)=frac{log(z)}{left (1+z^2 right )^2}$$
I consider the contour $gamma$ comprising two semi-circular arcs with radii $R$ and $epsilon$ (having redefined the logarithm by deleting the negative imaginary axis).
I'm having trouble with estimating the part of $gamma$ on the semi-circle $left | z right |=R$. I have:
$$left |int_{zin gamma ,left | z right |=R}f(z) right |leq int_{0}^{pi}left | f(Re^{itheta })iRe^{itheta } right |dthetaleq int_{0}^{pi}R frac{sqrt{log^{2}(R)+theta ^{2}}}{(R^{2}-1)^{2}}dtheta $$
My book does the estimation:
$$left | f(z) right |leq frac{2log(R)}{(R^{2}-1)^{2}}$$
How did they get the $2log(R)$ estimation?
Thank's in advance.
complex-analysis contour-integration
edited Dec 30 '18 at 14:14
mrtaurho
6,12251641
asked Dec 30 '18 at 13:55
John11John11
1,0421821
$endgroup$
2
$begingroup$
Assuming that $R$ is large enough we clearly have $theta^2 leq 3log^2(R)$ for any $thetain[0,pi]$.
$endgroup$
– Jack D'Aurizio
Dec 30 '18 at 14:14
$begingroup$
@JackD'Aurizio So I write that I take $R$ large enough to have that $log^2(R)geq frac{pi^2}{3}$ because $log^2(x)$ goes to infinity as x goes to infinity?
$endgroup$
– John11
Dec 30 '18 at 14:17
2
$begingroup$
I would get the numerator as $le pi+ln R$ which I suppose is $le 2ln R$ for large enough $R$.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 14:26
$begingroup$
@LordSharktheUnknown I like your idea as well! Thank you!
$endgroup$
– John11
Dec 30 '18 at 14:39
add a comment |
$begingroup$
I'm doing the classical example of treating the integral $int_{0}^{infty }frac{log(x)}{left (1+x^2 right )^2}dx$ using residue theorem on the function $$f(z)=frac{log(z)}{left (1+z^2 right )^2}$$
I consider the contour $gamma$ comprising two semi-circular arcs with radii $R$ and $epsilon$ (having redefined the logarithm by deleting the negative imaginary axis).
I'm having trouble with estimating the part of $gamma$ on the semi-circle $left | z right |=R$. I have:
$$left |int_{zin gamma ,left | z right |=R}f(z) right |leq int_{0}^{pi}left | f(Re^{itheta })iRe^{itheta } right |dthetaleq int_{0}^{pi}R frac{sqrt{log^{2}(R)+theta ^{2}}}{(R^{2}-1)^{2}}dtheta $$
My book does the estimation:
$$left | f(z) right |leq frac{2log(R)}{(R^{2}-1)^{2}}$$
How did they get the $2log(R)$ estimation?
Thank's in advance.
complex-analysis contour-integration
edited Dec 30 '18 at 14:14
mrtaurho
6,12251641
asked Dec 30 '18 at 13:55
John11John11
1,0421821
$endgroup$
I'm doing the classical example of treating the integral $int_{0}^{infty }frac{log(x)}{left (1+x^2 right )^2}dx$ using residue theorem on the function $$f(z)=frac{log(z)}{left (1+z^2 right )^2}$$
I consider the contour $gamma$ comprising two semi-circular arcs with radii $R$ and $epsilon$ (having redefined the logarithm by deleting the negative imaginary axis).
I'm having trouble with estimating the part of $gamma$ on the semi-circle $left | z right |=R$. I have:
$$left |int_{zin gamma ,left | z right |=R}f(z) right |leq int_{0}^{pi}left | f(Re^{itheta })iRe^{itheta } right |dthetaleq int_{0}^{pi}R frac{sqrt{log^{2}(R)+theta ^{2}}}{(R^{2}-1)^{2}}dtheta $$
My book does the estimation:
$$left | f(z) right |leq frac{2log(R)}{(R^{2}-1)^{2}}$$
How did they get the $2log(R)$ estimation?
Thank's in advance.
complex-analysis contour-integration
complex-analysis contour-integration
edited Dec 30 '18 at 14:14
mrtaurho
6,12251641
asked Dec 30 '18 at 13:55
John11John11
1,0421821
edited Dec 30 '18 at 14:14
mrtaurho
6,12251641
asked Dec 30 '18 at 13:55
John11John11
1,0421821
edited Dec 30 '18 at 14:14
mrtaurho
6,12251641
edited Dec 30 '18 at 14:14
mrtaurho
6,12251641
edited Dec 30 '18 at 14:14
mrtaurho
6,12251641
6,12251641
asked Dec 30 '18 at 13:55
John11John11
1,0421821
asked Dec 30 '18 at 13:55
John11John11
1,0421821
asked Dec 30 '18 at 13:55
John11John11
1,0421821
1,0421821
2
$begingroup$
Assuming that $R$ is large enough we clearly have $theta^2 leq 3log^2(R)$ for any $thetain[0,pi]$.
$endgroup$
– Jack D'Aurizio
Dec 30 '18 at 14:14
$begingroup$
@JackD'Aurizio So I write that I take $R$ large enough to have that $log^2(R)geq frac{pi^2}{3}$ because $log^2(x)$ goes to infinity as x goes to infinity?
$endgroup$
– John11
Dec 30 '18 at 14:17
2
$begingroup$
I would get the numerator as $le pi+ln R$ which I suppose is $le 2ln R$ for large enough $R$.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 14:26
$begingroup$
@LordSharktheUnknown I like your idea as well! Thank you!
$endgroup$
– John11
Dec 30 '18 at 14:39
add a comment |
2
$begingroup$
Assuming that $R$ is large enough we clearly have $theta^2 leq 3log^2(R)$ for any $thetain[0,pi]$.
$endgroup$
– Jack D'Aurizio
Dec 30 '18 at 14:14
$begingroup$
@JackD'Aurizio So I write that I take $R$ large enough to have that $log^2(R)geq frac{pi^2}{3}$ because $log^2(x)$ goes to infinity as x goes to infinity?
$endgroup$
– John11
Dec 30 '18 at 14:17
2
$begingroup$
I would get the numerator as $le pi+ln R$ which I suppose is $le 2ln R$ for large enough $R$.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 14:26
$begingroup$
@LordSharktheUnknown I like your idea as well! Thank you!
$endgroup$
– John11
Dec 30 '18 at 14:39
2
2
$begingroup$
Assuming that $R$ is large enough we clearly have $theta^2 leq 3log^2(R)$ for any $thetain[0,pi]$.
$endgroup$
– Jack D'Aurizio
Dec 30 '18 at 14:14
$begingroup$
Assuming that $R$ is large enough we clearly have $theta^2 leq 3log^2(R)$ for any $thetain[0,pi]$.
$endgroup$
– Jack D'Aurizio
Dec 30 '18 at 14:14
$begingroup$
@JackD'Aurizio So I write that I take $R$ large enough to have that $log^2(R)geq frac{pi^2}{3}$ because $log^2(x)$ goes to infinity as x goes to infinity?
$endgroup$
– John11
Dec 30 '18 at 14:17
$begingroup$
@JackD'Aurizio So I write that I take $R$ large enough to have that $log^2(R)geq frac{pi^2}{3}$ because $log^2(x)$ goes to infinity as x goes to infinity?
$endgroup$
– John11
Dec 30 '18 at 14:17
2
2
$begingroup$
I would get the numerator as $le pi+ln R$ which I suppose is $le 2ln R$ for large enough $R$.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 14:26
$begingroup$
I would get the numerator as $le pi+ln R$ which I suppose is $le 2ln R$ for large enough $R$.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 14:26
$begingroup$
@LordSharktheUnknown I like your idea as well! Thank you!
$endgroup$
– John11
Dec 30 '18 at 14:39
$begingroup$
@LordSharktheUnknown I like your idea as well! Thank you!
$endgroup$
– John11
Dec 30 '18 at 14:39
add a comment |
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2
$begingroup$
Assuming that $R$ is large enough we clearly have $theta^2 leq 3log^2(R)$ for any $thetain[0,pi]$.
$endgroup$
– Jack D'Aurizio
Dec 30 '18 at 14:14
$begingroup$
@JackD'Aurizio So I write that I take $R$ large enough to have that $log^2(R)geq frac{pi^2}{3}$ because $log^2(x)$ goes to infinity as x goes to infinity?
$endgroup$
– John11
Dec 30 '18 at 14:17
2
$begingroup$
I would get the numerator as $le pi+ln R$ which I suppose is $le 2ln R$ for large enough $R$.
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 14:26
$begingroup$
@LordSharktheUnknown I like your idea as well! Thank you!
$endgroup$
– John11
Dec 30 '18 at 14:39