- #1

- 116

- 0

## Homework Statement

[tex] \int^{\infty}_{0} \frac {\sqrt{x}}{1+x^{2}} dx [/tex]

## Homework Equations

## The Attempt at a Solution

[tex] f(z) = \frac {\sqrt{z}}{1+z^{2}} = \frac {\sqrt{z}}{(z-i)(z+i)} [/tex]

[tex] Res[f,i] = \lim_{z \to i} \frac {\sqrt {z}}{z+i}} = \frac {\sqrt {i}}{2i} = \frac {1}{2i} (\frac {\sqrt {2}}{2}} + i \frac {\sqrt {2}}{2}}) = \frac {\sqrt {2}}{4} - i \frac {\sqrt {2}}{4} [/tex]

[tex] Res[f,-i] = \lim_{z \to -i} \frac {\sqrt {z}}{z-i}} = \frac {\sqrt {-i}}{-2i} = \frac {1}{-2i} (\frac {\sqrt {2}}{2}} - i \frac {\sqrt {2}}{2}}) = \frac {\sqrt {2}}{4} + i \frac {\sqrt {2}}{4} [/tex]

[tex] \int^{\infty}_{0} \frac {\sqrt{x}}{1+x^{2}} dx = \frac {2 \pi i}{1-e^{i \pi}} [ \frac {\sqrt {2}}{4} - i \frac {\sqrt {2}}{4} + \frac {\sqrt {2}}{4} + i \frac {\sqrt {2}}{4}] = \frac {2 \pi i}{2}( \frac {\sqrt {2}}{2}) = i \frac {\sqrt {2}}{2} \pi [/tex]

I've done this problem over and over again and I can't figure out where I made a mistake. The answer obviously can't be imaginary.