WLOG we assume that \(0<a<1\). For the calculation of \
we consider the complex function \(f(z):=\dfrac{z^a}{z^2-z\sqrt{2}+1}=\dfrac{{\rm{e}}^{a\,{\rm{Log}\,z}}}{z^2-z\sqrt{2}+1}\,, \; z\in{\mathbb{C}}\setminus\{x+0i\;|\;x<0\}\,,\) and the simple, closed path \(\gamma\) which is the summand of the anticlockwise oriented arc \(C_r:=\big\{r\,{\rm{e}}^{it}\in{\mathbb{C}}\;\big|\; t\in(-\pi+\varepsilon,\pi-\varepsilon]\big\}\,, \; r\to+\infty\,,\;\varepsilon\to0^+\,, \) of the segment \(\ell:=\big\{(r+\delta-t)\,{\rm{e}}^{i\pi}\in{\mathbb{C}}\;\big|\; t\in(\delta,r)\big\}\), of the clockwise oriented arc \(C_\delta:=\big\{\delta\,{\rm{e}}^{-it}\in{\mathbb{C}}\;\big|\; t\in(-\pi+\varepsilon,\pi-\varepsilon]\big\}\,, \;\delta\to0^{+}\,,\;\varepsilon\to0^+\,,\) and of the segment \(-\ell:=\big\{t\,{\rm{e}}^{-i\pi}\in{\mathbb{C}}\;\big|\; t\in(\delta,r)\big\}\).
[/centre]
The function \(f\) is meromorphic in \({\mathbb{C}}\setminus\{x+0i\;|\;x<0\}\) with simple poles \(z_1=\frac{\sqrt{2}}{2}(1+i)\) and \(z_2=\frac{\sqrt{2}}{2}(1-i)\). So, we have
\begin{align*}
{\rm{Res}\,}\big(f(z), z=\tfrac{\sqrt{2}}{2}(1+i)\big)&=\frac{1}{0!}\,\mathop{\lim}\limits_{z\to\frac{\sqrt{2}}{2}(1+i)}\bigg(\big( z-\tfrac{\sqrt{2}}{2}(1+i)\big)\dfrac{{\rm{e}}^{a{\rm{Log}\,z}}}{\big( z-\tfrac{\sqrt{2}}{2}(1+i)\big)\big( z-\tfrac{\sqrt{2}}{2}(1-i)\big)}\bigg)\\
&=\mathop{\lim}\limits_{z\to\frac{\sqrt{2}}{2}(1+i)}\dfrac{{\rm{e}}^{a{\rm{Log}\,z}}}{z-\tfrac{\sqrt{2}}{2}(1-i)}\\
&= -\frac{i\sqrt{2}}{2}\,{\rm{e}}^{\frac{\pi a}{4}i}\,,\\
{\rm{Res}\,}\big(f(z), z=\tfrac{\sqrt{2}}{2}(1-i)\big)&=\frac{1}{0!}\,\mathop{\lim}\limits_{z\to\frac{\sqrt{2}}{2}(1-i)}\bigg(\big( z-\tfrac{\sqrt{2}}{2}(1-i)\big)\dfrac{{\rm{e}}^{a{\rm{Log}\,z}}}{\big( z-\tfrac{\sqrt{2}}{2}(1+i)\big)\big( z-\tfrac{\sqrt{2}}{2}(1-i)\big)}\bigg)\\
&=\mathop{\lim}\limits_{z\to\frac{\sqrt{2}}{2}(1-i)}\dfrac{{\rm{e}}^{a{\rm{Log}\,z}}}{z-\tfrac{\sqrt{2}}{2}(1+i)}\\
&=\frac{i\sqrt{2}}{2}\,{\rm{e}}^{-\frac{\pi a}{4}i}
\end{align*}
and
\begin{align*}
\displaystyle\oint_{\gamma}{f(z)\,dz}&=2\pi i\;\cancelto{1}{{\rm{I}\,}\big(\gamma, z=\tfrac{\sqrt{2}}{2}(1+i)\big)}\,{\rm{Res}\,}\big(f(z), z=\tfrac{\sqrt{2}}{2}(1+i)\big)\,+\\
&\qquad2\pi i\;\cancelto{1}{{\rm{I}\,}\big(\gamma, z=\tfrac{\sqrt{2}}{2}(1-i)\big)}\,{\rm{Res}\,}\big(f(z), z=\tfrac{\sqrt{2}}{2}(1-i)\big)\\
&=-2\pi i\,\frac{i\sqrt{2}}{2}\,{\rm{e}}^{\frac{\pi a}{4}i}+2\pi i\,\frac{i\sqrt{2}}{2}\,{\rm{e}}^{-\frac{\pi a}{4}i}\\
&=\pi\sqrt{2}\,\big({\rm{e}}^{\frac{\pi a}{4}i}-{\rm{e}}^{-\frac{\pi a}{4}i}\big)\\
&=\pi\sqrt{2}\;2i\sin\big(\tfrac{\pi a}{4}\big)&\Longrightarrow\\
\displaystyle\oint_{C_{r}}{f(z)\,dz}+\oint_{\ell}{f(z)\,dz}&+\oint_{C_{\delta}}{f(z)\,dz}+\oint_{-\ell}{f(z)\,dz}=2\pi\,i\sqrt{2}\,\sin\big(\tfrac{\pi a}{4}\big)\qquad{(1)}\,.
\end{align*} By Estimation lemma we have that \[\displaystyle\mathop{\lim}\limits_{r\to+\infty}\oint_{C_{r}}{f(z)\,dz}=\mathop{\lim}\limits_{\delta\to0^{+}}\oint_{C_{\delta}}{f(z)\,dz}=0\,.\]
Also \begin{align*}
\displaystyle\oint_{\ell}{f(z)\,dz}&=\int_{\delta}^{r}{\dfrac{|r+\delta-t|^a\,{\rm{e}}^{a\pi i}}{(r+\delta-t)^2{\rm{e}}^{2\pi i}-(r+\delta-t)\,{\rm{e}}^{\pi i}\sqrt{2}+1}\,dt}\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{s\,=\,r+\delta-t}\\
{ds\,=\,-dt}\\
\end{subarray}}\,-\int_{r}^{\delta}{\dfrac{|s|^a\,{\rm{e}}^{a\pi i}}{s^2+s\sqrt{2}+1}\,ds}\\
&=\int_{\delta}^{r}{\dfrac{t^a\,{\rm{e}}^{a\pi i}}{t^2+t\sqrt{2}+1}\,dt}\,,\\
\displaystyle\oint_{-\ell}{f(z)\,dz}&=\int_{\delta}^{r}{\dfrac{|t|^a\,{\rm{e}}^{-a\pi i}}{t^2{\rm{e}}^{2\pi i}-t{\rm{e}}^{\pi i}\sqrt{2}+1}\,(-1)\,dt}\\
&=-\int_{\delta}^{r}{\dfrac{t^a\,{\rm{e}}^{a\pi i}}{t^2+t\sqrt{2}+1}\,dt}\qquad\qquad\qquad\qquad\qquad\Longrightarrow\\
\oint_{\ell}{f(z)\,dz}+\oint_{-\ell}{f(z)\,dz}&=\int_{\delta}^{r}{\dfrac{t^a\,{\rm{e}}^{a\pi i}}{t^2+t\sqrt{2}+1}\,dt}-\int_{\delta}^{r}{\dfrac{t^a\,{\rm{e}}^{-a\pi i}}{t^2+t\sqrt{2}+1}\,dt}\\
&=\big({\rm{e}}^{a\pi i}-{\rm{e}}^{-a\pi i}\big)\int_{\delta}^{r}{\dfrac{t^a}{t^2+t\sqrt{2}+1}\,dt}\\
&=2i\sin(\pi a)\int_{\delta}^{r}{\dfrac{t^a}{t^2+t\sqrt{2}+1}\,dt}\end{align*}
Using the above calculations the equation $(1)$ becames \begin{align*}
2i\sin(\pi a)\mathop{\lim}\limits_{\begin{subarray}{c}
{r\to+\infty}\\
{\delta\to0^+}
\end{subarray}}\int_{\delta}^{r}{\dfrac{t^a}{t^2+t\sqrt{2}+1}\,dt}\,+\\
\cancelto{0}{\mathop{\lim}\limits_{\begin{subarray}{c}
{r\to+\infty}\\
{\delta\to0^+}
\end{subarray}}\int_{C_{r}}{f(z)\,dz}}+\cancelto{0}{\Bigg.\mathop{\lim}\limits_{\begin{subarray}{c}
{r\to+\infty}\\
{\delta\to0^+}
\end{subarray}}\int_{C_{\delta}}{f(z)\,dz}}&\stackrel{(1)}{=}2\pi\,i\sqrt{2}\,\sin\big(\tfrac{\pi a}{4}\big)\qquad\qquad \Longrightarrow\\
\mathop{\lim}\limits_{\begin{subarray}{c}
{r\to+\infty}\\
{\delta\to0^+}
\end{subarray}}\int_{\delta}^{r}{\dfrac{t^a}{t^2+t\sqrt{2}+1}\,dt}&=\pi\sqrt{2}\,\frac{\sin\big(\tfrac{\pi a}{4}\big)}{\sin(\pi a)}\qquad\qquad \quad\Longrightarrow\\
\displaystyle\int_{0}^{+\infty}{\dfrac{t^a}{t^2+t\sqrt{2}+1}\,dt}&=\pi\sqrt{2}\,\frac{\sin\big(\tfrac{\pi a}{4}\big)}{\sin(\pi a)}\,.
\end{align*}