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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.