# Residue Calculus integrate Sqrt(x)ln(x)/(1+x^2)

## Homework Statement

Integrate from zero to infinity;

f(x)=\sqrt(x)log(x)/(x^2+1)

## Homework Equations

Branch cut makes log(z)= ln|z|+i Arg(z)

Poles are at +/- i and Res(z=i) is \pi/4 e^(i \pi/4)

I'll need to close the contour; probably as an annullus in the top half of the plane with r<1, R>1

## The Attempt at a Solution

I know that the total integral is 2\pi i Res(z=i) since that's the only pole enclosed in the annulus; but I can't see how to show that the semicircles go to zero as the radius goes to zero/infinity
I worked out that (converting z=r' e^ix
|∫f(z) dz|<=∫|fz|dz and |f(z)|= \sqrt((r')(ln|r'|^2+x^2))/(\sqrt(r'^2 cos^2(2x)+1)^2+r'^4 sin^2(2x))
|f(z)|<=\sqrt((r')(ln|r'|^2+x^2))/(r'^2) (by taking out the +1 in the denominator
But I don't see how this has a limit at zero as r->0 and R->\infty
Help would be much appreciated.

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## Homework Statement

Integrate from zero to infinity;

f(x)=\sqrt(x)log(x)/(x^2+1)

## Homework Equations

Branch cut makes log(z)= ln|z|+i Arg(z)

Poles are at +/- i and Res(z=i) is \pi/4 e^(i \pi/4)

I'll need to close the contour; probably as an annullus in the top half of the plane with r<1, R>1

## The Attempt at a Solution

I know that the total integral is 2\pi i Res(z=i) since that's the only pole enclosed in the annulus; but I can't see how to show that the semicircles go to zero as the radius goes to zero/infinity
I worked out that (converting z=r' e^ix
|∫f(z) dz|<=∫|fz|dz and |f(z)|= \sqrt((r')(ln|r'|^2+x^2))/(\sqrt(r'^2 cos^2(2x)+1)^2+r'^4 sin^2(2x))
|f(z)|<=\sqrt((r')(ln|r'|^2+x^2))/(r'^2) (by taking out the +1 in the denominator
But I don't see how this has a limit at zero as r->0 and R->\infty
Help would be much appreciated.
First consider the indentation around the origin. If we let $z=\rho e^{it}$, then the integral becomes:

$$\lim_{\rho\to 0} \int_a^b \frac{\rho^{1/2} e^{it/2} \left(\ln(\rho)+it\right) \rho i e^{it}}{\rho^2 e^{2 i t}+1} dt$$

Now, what is:

$$\lim_{\rho\to 0} \frac{\rho^{1/2} e^{it/2} \left(\ln(\rho)+it\right) \rho i e^{it}}{\rho^2 e^{2 i t}+1}$$

Well surely the denominator goes to just one right? Scrap everything in the numerator that's not a $\rho$ (we could just let them equal a big number for that matter and it won't affect the limit). And so you're left with:

$$\lim_{\rho\to 0} \rho^{3/2} \ln(\rho)$$

Now you try and do the same type of analysis with $R\to\infty$ for the larger arc.

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Thanks thatreally helped. i got it out now