Calculating an integral threw residium question

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i need to calculate this integral f(z)=\frac{z}{e^{2\pi iz^2}-1}\\

in this area
\gamma _r=\left \{ |z|=r \right \},r>2

i need to find the points which turn to zero in the denominator
and non zero in the numerator.
i got two such points
z=\pm \sqrt{n}
by using this formula
res(\sqrt{a})=\frac{p(a)}{q(a)'}
res(\sqrt{n})=\frac{1}{4\pi i}
res(-\sqrt{n})=\frac{1}{4\pi i}

the third point is z=0 but for it we have both numerator and denominator 0
i calculated the residium for it by res(f(x),a)=\lim_{x->a}(f(x)(x-a)) formula
but then
my prof says some stuff that involves the area
he says that my points are 0 +1 -1 +2^(0.5) -2^(0.5) etc.. because the denominator goes to zero
for each point have a residiu and i need to sum the residiums inside.
but here the area is not defined
its not like (by radius 3)

i don't know what point are inside the area

??
 
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anyone?>?>
 
sorry there is a mistake
the area is
<br /> \gamma _r=\left \{ |z|=r \right \},n&lt;r^2&lt;n+1<br />

and the integral is from plus to minus infinity
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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