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I tried to find the partial fractions of $\frac{1}{(z^2+1)^2}$ then

$\frac{1}{(z^2+1)^2}=\frac{A}{(z+i)^2}+\frac{B}{(z+i)}+\frac{C}{(z-i)^2}+\frac{D}{(z-i)}$

$1=A(z-i)^2+B(z+i)(z-i)^{2}+C(z+i)^2+D(z+i)^2(z-i)$

I got $A=D=-\frac{1}{4}$ but I am not sure how to get $B$ and $C$?

I there a way to find the values of $B$ and $C$? Also, is there a way to solve this integral without finding the partial fractions?

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    You can foil it all out. – 2017-01-07
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    @SimpleArt, what do you mean by $foil it all out"? – 2017-01-07
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    what is the purpose of this strange partial fractions decomposition? – 2017-01-07
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    One can use the residue theorem: write $\frac{1}{(z^2+1)^2} = \frac{1}{(z-i)^2(z+i)^2}$ which gives you the poles. The residue of a pole of order $2$ at $z=z_0$ is given by $\frac{d}{dz}\left[(z-z_0)^2f(z)\right]_{z=z_0}$. Both of the poles are inside the contour. – 2017-01-07
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    @Winther, what is $z_0$ in this case? – 2017-01-07
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    @gbd $z_0$ is the pole in question. So using $z_0 = i$ and $z_0 = -i$ will give the two residues needed to solve this problem. – 2017-01-07
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    @Winther, is the answer then given by $f^{\prime}(i)+f^{\prime}(-i)$? – 2017-01-07
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    In this case $f(z) = \frac{e^{zt}}{(z^2+1)^2}$ so for example for the pole at $z=i$ we have: $\text{Res}[\frac{e^{zt}}{(z^2+1)^2};z=i] = \lim_{z\to i} \frac{d}{dz}\left[(z-i)^2\frac{e^{zt}}{(z^2+1)^2}\right] = \lim_{z\to i} \frac{d}{dz}\left[\frac{e^{zt}}{(z+i)^2}\right] = \lim_{z\to i}\left[\frac{te^{zt}}{(z+i)^2} - \frac{2te^{zt}}{(z+i)^3}\right]$. Then you also need the other residue at $z=-i$ computed similary to get $\oint f(z){\rm d}z = 2\pi i[\text{Res}[\frac{e^{zt}}{(z^2+1)^2};z=i]+\text{Res}[\frac{e^{zt}}{(z^2+1)^2};z=-i]]$. – 2017-01-07
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    Is that $\;e^{zt}\;$ ? What is $\;t\;$ here? – 2017-01-07

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Hint:

Letting $\;C_1,\,C_2\;$ being little circle around the poles $\;i,\,-i\;$ resp., we have that

$$\frac1{2\pi i}\oint_{|z|=3}\frac{e^{zt}}{(z^2+1)^2}dz=\frac1{2\pi i}\left[\oint_{C_1}\frac{\frac{e^{zt}}{(z+i)^2}}{(z-i)^2}dz+\oint_{C_2}\frac{\frac{e^{zt}}{(z-i)^2}}{(z+i)^2}dz\right]=$$

$$=\left.\left(\frac{e^{zt}}{(z+i)^2}\right)'\right|_{z=i}+\left.\left(\frac{e^{zt}}{(z-i)^2}\right)'\right|_{z=-i}$$

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    Why not appeal directly to the residue theorem? I fear the detour of introducing "little circles" is just going to cause confusion. – 2017-01-07
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    @Winther I think this method is way easier and more direct: using Cauchy's Integral Formula for the (first, in this case) derivative of an analytic function around each singularity of the whole function. – 2017-01-07
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    Way easier for someone just starting with residue calculus? I doubt it. It begs the questions for which there are no explanation here: why can I do this, can I always do it? Where does $|z| = 3$ enter in, can I do this if $|z| = 1/2$? Appealing directly to the residue theorem avoids all this magic of deforming the contour and it's much more straight forward (just sum residues inside the contour) which leaves little room for misunderstandings. – 2017-01-07
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    @Winther I already wrote down my contribution. You're invited to write down yours, and let the OP choose. And any question the OP may have he can ask if he wants to. – 2017-01-07