Exercises 7.41 and 8.14 from Rubinstein's book
Polymer Physics.
From Boltzmann superposition principle, we have
$$\sigma(t)=\int_{-\infty}^t G(t-t^\prime)\dot{\gamma}(t^\prime)\mathrm{d}t^\prime$$
$$ \gamma(t) = \gamma_0 \exp(I \omega t)$$
$$\begin{align} \sigma(t)&=I\omega \gamma_0 \int_{-\infty}^t G(t-t^\prime)\exp(I \omega t^\prime)\mathrm{d}t^\prime\\ &=I \omega \gamma_0\exp(I \omega t)\int_0^\infty G(\tau)\exp(-I\omega \tau)\mathrm{d}\tau \\ &:=\gamma(t)I \omega \hat{G}(\omega)\end{align}$$
We therefore have
$$I\omega\hat{G}=I\omega \mathcal{F}^{\cos}\lbrace G\rbrace+\omega \mathcal{F}^{\sin}\lbrace G\rbrace\equiv I G^{\prime\prime}+G^\prime$$
$\mathcal{F}^{\cos}$ and $\mathcal{F}^{\sin}$ represent
Fourier cosine and sine transform. Take $\mathcal{F}^{\cos}(1)=\delta(\omega)$ and $\mathcal{F}^{\sin}(1)=1/\omega$ (note the normalization factor of $\mathcal{F}$), let $G(t)=G_{eq}+G_1(t)$:
$$\begin{align} G^\prime &= \omega\mathcal{F}^{\sin}(G_1+G_{eq}) = G_{eq}+\omega\mathcal{F}^{\sin}(G_1(t))\\G^{\prime\prime}&=\omega\mathcal{F}^{\cos}(G_1+G_{eq})=\omega\mathcal{F}^{\cos}(G_1(t))+\omega\delta(\omega)=\omega\mathcal{F}^{\cos}(G_1(t))\end{align}$$
$G'$ and $G''$ are defined as the Fourier sine and cosine transform of stress relaxation function $G(t)$:
$$ \begin{align} G^\prime &:= \omega\int_0^\infty G(t) \sin\omega t\mathrm{d}t\\ G^{\prime\prime} &:= \omega\int_0^\infty G(t) \cos\omega t\mathrm{d}t \end{align}$$
Let $G(t)$ be
$$G(t)\sim \left(\frac{t}{\tau_0}\right)^{-\frac{1}{2}} \exp\left(-\frac{t}{\tau_R}\right)$$
G' i.e., the cosine transform is thus
$$G^{\prime\prime}/\omega=\int_0^\infty \mathrm{d}t \left(\frac{t}{\tau_0}\right)^{-\frac{1}{2}} \exp\left(-\frac{t}{\tau_R}\right)\cos\omega t$$
Since $\tau_R = N^2\tau_0$, we have (let ${t^\prime}^2\tau_R = t$)
$$\begin{align} G^{\prime\prime}/\omega &= 2\frac{\tau_R}{N} \int_0^\infty \mathrm{d}t^\prime \exp\left(-{t^\prime}^2\right)\cos\omega\tau_R {t^\prime}^2\\ &=2\frac{\tau_R}{N} \mathrm{Re} \int_0^\infty \mathrm{d}t^\prime \exp\left(-{t^\prime}^2\right)\exp\left(I \omega \tau_R {t^\prime}^2\right)\\ &=2\frac{\tau_R}{N} \mathrm{Re} \int_0^\infty \mathrm{d}t^\prime \exp\left(-(1-I \omega \tau_R){t^\prime}^2\right) \end{align} $$
Since $t^2=(-t)^2$, extending the range of last integral to $(-\infty, \infty)$ yields a
Gaussian integral with complex argument, the result is straightforward:
$$ \begin{align}G^{\prime\prime}/\omega &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\frac{\pi}{1-I \omega \tau_R}}\\ &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\pi}\frac{\sqrt{1+I\omega\tau_R}}{1+(\omega\tau_R)^2}\\ &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\pi}\frac{1}{1+(\omega\tau_R)^2}\sqrt{\sqrt{1+(\omega\tau_R)^2}}\times&\\ &\left(\cos\left(\frac{1}{2}\arctan\omega\tau_R\right)+I \sin\left(\frac{1}{2}\arctan\omega\tau_R\right)\right)\\ &= \frac{\tau_R}{N}\sqrt{\pi}\frac{1}{1+(\omega\tau_R)^2}\sqrt{\sqrt{1+(\omega\tau_R)^2}}\cos\left(\frac{1}{2}\arctan\omega\tau_R\right)\end{align} $$
With $\cos(\arctan(x)) = \frac{1}{1+x^2}$ and $2\cos(x/2)^2-1=\cos(x)$, we finally have
$$G^{\prime\prime} = \omega\tau_R\sqrt{\frac{\pi}{2}} \sqrt{\frac{\sqrt{1+(\omega\tau_R)^2}+1}{1+(\omega\tau_R)^2}}$$
The factor $\frac{\pi}{2}$ comes from the definition of Fourier cosine transform.
P.S. It is more convinient (non-restrictively) to treat the integral
$$t^{-\alpha} \exp(-t) \exp(I w t)$$
with $0\lt \alpha \lt 1$ as Gamma function with (complex) coefficient in $\exp$:
$$\int_0^\infty \mathrm{d}t t^{-\alpha}\exp(-(1-I \omega) t) = (1-I\omega t)^{\alpha-1} \Gamma(1-\alpha)$$
We may use Hypergeometric function $_pF_q$ for more genernal cases, i.e., $t^{-\alpha}\exp(-t^\beta)\exp(I w t)$
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