I have an equation derived from Ising model for the correlation distance. Is there a way to rewrite this integral to make in manageable?

For $d=3$ we have for $t>0$

$$\begin{align}C(\vec r)&=\int_0^{2\pi}\int_0^\pi \int_0^\infty \frac{e^{ikr\cos(\theta)}}{t+k^2}\,k^2\sin(\theta)\,dk\,d\theta\,d\phi\\\\ &=\frac{2\pi}r \int_{-\infty}^\infty \frac{k\sin(kr)}{t+k^2}\,dk\\\\ &=\frac{2\pi}{r}\text{Im}\left(\text{PV}\int_{-\infty}^\infty \frac{ke^{ikr}}{t+k^2}\,dk\right)\\\\ &=\frac{2\pi}{r} \text{Im}\left(2\pi i \frac{i\sqrt{t}e^{-r\sqrt{t}}}{2i\sqrt{t}}\right)\\\\ &=2\pi^2\frac{ e^{-\sqrt{t} r}}{r} \end{align}$$

Can you generlize this approach using mathematics of the $n$-Sphere?

One way to find the solution for $d=n$ is to invoke the theory of distributions. Note that in distribution, we have

$$(\Delta -t)C(\vec r)=-(2\pi)^n\delta(\vec r)$$

Then, using the result I developed in THIS ANSWER with $k\mapsto i\sqrt{t}$ and assuming a time convention $e^{-i\omega \tau}$, we find

$$C(\vec r)=(2\pi)^n\frac i4\left(\frac{i\sqrt{t}}{2\pi r}\right)^{n/2-1}H_{n/2-1}^{(1)}(i\sqrt{t}r)$$

If we wish to find the first term in the large $r$ asymptotic expanion of $C(\vec r)$, we simply use the well-known large argument asymptotic relationship

$$H_{n-1/2}^{(1)}(i\sqrt{t}r)\sim\sqrt{\frac{2}{\pi i\sqrt{t}r}}e^{-in\pi/2}e^{-\sqrt{t}r}$$

I'll leave the arithmetic to the reader.

There is a nice solution by @Mark Viola for $n=3$ - which can be generalize for higher $n$. To find the asymptotics for $r\to\infty$, we can also use the Laplace' method.

Let $a^2=t$, and

$$I(n,a)=\int d^n k { e^{i(\vec k \cdot \vec{r})}\over a^2+ k^2}=\int d^n k e^{i(\vec k \cdot \vec{r})}\int_0^\infty e^{-x (a^2+ k^2)}dx$$ Changing the order of integration $$I(n,a)=\int_0^\infty e^{-x a^2}dx\int_{-\infty}^\infty..\int_{-\infty}^\infty dk_1...dk_n e^{-x (k^2_1+...+k^2_n)}e^{i(k_1r_1+...+k_nr_n)}$$ Using $$-xk^2_l+ik_lr_l=-x\Big(k^2_l-\frac{ir_l k_l}{x}-\frac{r^2_l}{4x^2}+\frac{r^2_l}{4x^2}\Big)=-x\Big(k_l-\frac{ir_l}{2}\Big)^2-\frac{r^2_l}{4x}$$ $$I(n,a)=\int_0^\infty e^{-x a^2}dx\bigg(\int_{-\infty}^\infty dk_le^{-x\Big(k_l-\frac{ir_l}{2}\Big)^2}\bigg)^ne^{-\frac{(r^2_1+...+r^2_n)}{4x}}$$ $$=\int_0^\infty e^{-x a^2}e^{-\frac{r^2}{4x}}\Big(\frac{\pi}{x}\Big)^{\frac{n}{2}} dx=\int_0^\infty \Big(\frac{\pi}{x}\Big)^{\frac{n}{2}}e^{-f(x)} dx$$ where $f(x)=xa^2+\frac{r^2}{4x}$. To find the asymptotics at $r\to\infty$ we use the Laplace' method. $f'(x)=0$ at $x=x_0=\frac{r}{2a}$; $f\big(\frac{r}{2a}\big)=ar; f''\big(\frac{r}{2a}\big)=\frac{4a^3}{r}$. Using the expansion of $f(x)$ near this critical point $(f(x)=f(x_0)+\frac{1}{2}f''(x_0)(x-x_0)^2+...$), and extending the integration to $\pm\infty$, we find the main asymptotics term: $$I(n,a)\sim\Big(\frac{2\pi a}{r}\Big)^{\frac{n}{2}} e^{-ar}\int_{-\infty}^\infty e^{-\frac{2 a^3}{r}(x-x_0)^2}dx=\Big(\frac{2\pi a}{r}\Big)^{\frac{n}{2}}\sqrt\frac{\pi r}{2a^3}e^{-ar}$$ $$I(n,a)\sim\frac{(2\pi)^{\frac{n+1}{2}}}{2a}\Big(\frac{a}{r}\Big)^{\frac{n-1}{2}}e^{-ra}$$ For $n=3$ we get the exact result, obtained by @Mark Viola. The answer can also be obtained by means of the direct integration in this specific case (making the change of the variable $y=\frac{r}{2\sqrt x}$ and applying the Glasser's master theorem).