Integral using Gamma and Beta function

Question

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

$$\int_{\mathbb{R}^n}(a+|x|)^m\exp\left(-\lambda\left[\sum_{i=1}^nx_i^2\right]^k\right)d^nx$$ $$x=\mathbb{R}^n,k>0,m\in\mathbb{N},\Re(\lambda)>0$$

Answer 1

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 :)

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

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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

Answer 2

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 $$ \omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\tag5 $$

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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