Integral using Gamma and Beta function the integral I have found this integral on past exams papers for a course in Mathematical Methods of Physics. I really cant find any material on the textbook which could be helpfull for the solution. Do you have any idea how to solve this or can you suggest any material I could use? I post the question in the picture bellow. I have to express the solution using Beta and Gamma functions. Any help would be valuable. Thanks in advance!
Thanks for the edit, its the 1st time I post here, sorry if my post was wrong
$$\int_{\mathbb{R}^n}(a+|x|)^m\exp\left(-\lambda\left[\sum_{i=1}^nx_i^2\right]^k\right)d^nx$$
$$x=\mathbb{R}^n,k>0,m\in\mathbb{N},\Re(\lambda)>0$$
 A: Deriving the Surface Area of a Sphere
Start with the $1$ dimensional integral
$$
\int_{-\infty}^\infty e^{-\pi x^2}\,\mathrm{d}x=1\tag1
$$
and take the product of $n$ copies:
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty e^{-\pi \left(x_1^2+x_2^2+\cdots+x_n^2\right)}\,\mathrm{d}x_1\,\mathrm{d}x_2\cdots\,\mathrm{d}x_n=1\tag2
$$
Since the integrand is constant over spherical shells, we can easily write it as a spherical integral
$$
\int_0^\infty e^{-\pi r^2}\omega_{n-1}r^{n-1}\,\mathrm{d}r=1\tag3
$$
where $\omega_{n-1}$ is the surface area of the $n-1$ dimensional unit sphere.
Thus, we get
$$
\begin{align}
1
&=\int_0^\infty e^{-\pi r^2}\omega_{n-1}r^{n-1}\,\mathrm{d}r\tag{4a}\\
&=\frac{\omega_{n-1}}2\int_0^\infty e^{-\pi r^2}r^{n-2}\,\mathrm{d}r^2\tag{4b}\\
&=\frac{\omega_{n-1}}2\int_0^\infty e^{-\pi r}r^{n/2-1}\,\mathrm{d}r\tag{4c}\\
&=\frac{\omega_{n-1}}{2\pi^{n/2}}\int_0^\infty e^{-r}r^{n/2-1}\,\mathrm{d}r\tag{4d}\\
&=\frac{\omega_{n-1}}{2\pi^{n/2}}\Gamma(n/2)\tag{4e}
\end{align}
$$
Explanation:
$\text{(4a)}$: copy $(3)$
$\text{(4b)}$: $r\,\mathrm{d}r=\frac12\mathrm{d}r^2$ and pull the constant out front
$\text{(4c)}$: substitute $r\mapsto r^{1/2}$
$\text{(4d)}$: substitute $r\mapsto r/\pi$
$\text{(4e)}$: definition of the Gamma function
Therefore, we get
$$
\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\tag5
$$

Application to the Question
We can write the integral from the question as an integral on spherical shells:
$$
\begin{align}
&\int_0^\infty\overbrace{(a+r)^me^{-\lambda r^{2k}}}^{\substack{\text{integrand on a}\\\text{shell of radius $r$}}}\overbrace{\omega_{n-1}r^{n-1\vphantom{r^2}}}^{\substack{\text{surface area}\vphantom{g}\\\text{of the shell}}}\,\mathrm{d}r\tag{6a}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}\int_0^\infty a^{m-j}r^{\frac{j+n}{2k}-1}e^{-\lambda r}\,\mathrm{d}r\tag{6b}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}a^{m-j}\lambda^{-\frac{j+n}{2k}}\int_0^\infty r^{\frac{j+n}{2k}-1}e^{-r}\,\mathrm{d}r\tag{6c}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}a^{m-j}\lambda^{-\frac{j+n}{2k}}\Gamma\!\left(\frac{j+n}{2k}\right)\tag{6d}
\end{align}
$$
Explanation:
$\text{(6a)}$: integral from the question
$\text{(6b)}$: substitute $r\mapsto r^{\frac1{2k}}$ and apply the Binomial Theorem
$\text{(6c)}$: substitute $r\mapsto r/\lambda$
$\text{(6d)}$: definition of the Gamma function
A: I think one thing you can do is use hyperspherical coordinates i.e:
$$r^2=\sum_{i=1}^nx_i^2$$
making:
$$\exp\left(-\lambda\left[\sum_{i=1}^nx_i^2\right]^k\right)=\exp\left(-\lambda r^{2k}\right)$$
$$\left(a+|x|\right)^m=(a+r)^m$$
and now we have have to use the Jacobian to convert from cartesian to hyperspherical:
$$d^nx=r^{n-1}dr\prod_{i=1}^{n-1}\sin^{i-1}(\varphi_i)d\varphi_i$$
now in terms of our domain for each of these, we have:
$$r\in[0,\infty),\,\varphi_{1}\in[0,2\pi),\varphi_{i\ne1}\in[0,\pi]$$
so now if we reconstruct our integral:
$$I_n=\int_{r=0}^\infty\int_{\varphi_1=0}^{2\pi}\int_{\varphi_{i\ne1}\in[0,\pi]^{n-2}}r^{n-1}(a+r)^m\exp\left(-\lambda r^{2k}\right)dr\prod_{i=1}^{n-1}\sin^{i-1}(\varphi_i)d\varphi_i$$
now we can split this up as the function is separable, and we can write it as:
$$I=\left(\int_{r=0}^\infty r^{n-1}(a+r)^m\exp(-\lambda r^{2k})dr\right)\left(\int_{\varphi_1=0}^{2\pi}d\varphi_1\right)\prod_{i=2}^{n-1}\left(\int_{\varphi_i=0}^\pi\sin^{i-1}(\varphi_i)d\varphi_i\right)$$
Now the rest should be fairly easy :)

Note that:
$$\int_0^\pi\sin^{i-1}(\alpha)d\alpha=B\left(\frac12,\frac i2\right)$$
so your integral can be simplified to:
$$I=2\pi\prod_{i=2}^{n-1}B\left(\frac12,\frac n2\right)\int_0^\infty r^{n-1}(a+r)^m\exp(-\lambda r^{2k})dr$$

I think this integral could get quite messy due to the power in the exponential, so the next step I would take is use the fact that:
$$\exp(-\lambda r^{2k})=\sum_{j=0}^\infty(-1)^j\frac{\lambda^j r^{2jk}}{j!}$$
then the integral we have left could be represented in terms of the incomplete beta function
