$$ \int_0^1 \! \int_0^1 ...\int_0^1 \frac{dx_1dx_2...dx_n}{(a+x_1+x_2+...+x_n)^m}$$
I've tried for
n=2, m=1, I got : $a\ln a - 2(a+1)\ln (a+1)+(a+2)\ln(a+2)$
n=2, m=2, I got : $-\ln a + 2\ln(a+1)-\ln(a+2)$
n=3, m=1, I got : $\frac{1}{2}(-a^2\ln a+3(a+1)^2\ln (a+1)-3(a+2)^2\ln(a+2)+(a+3)^2\ln (a+3))$
but I still have no idea what's next. Or there is a tricky way??
On
Not a Complete Answer
For $n=4$, $m=1$ Mathematica gave me a similar answer:
$$\frac{1}{6}\left(a^3\ln a-4(a+1)^3\ln(a+1)+6(a+2)^3\ln(a+2)-4(a+3)^3\ln(a+3)+(a+4)^3\ln(a+4)\right)$$
And for $n=5$, $m=1$ it gave me:
$$\frac{1}{24}\left(-a^4\ln a+5(a+1)^4\ln(a+1)-10(a+2)^4\ln(a+2)+10(a+3)^4\ln(a+3)+-5(a+4)^4\ln(a+4)+(a+5)^4\ln(a+3)\right)$$
So for the $m=1$ family it may be reasonable to conclude that the answer is:
$$\frac{1}{(n-1)!}\sum_{i=0}^n(-1)^{n+i}{n \choose i}(a+i)^n\ln(a+i)$$
On
Solution
Here is a closed form solution of the multiple integral
$$f(n,m,a)=\frac{\Gamma (m-n)}{\Gamma (m)} \sum _{k=0}^n (-1)^k \binom{n}{k} (a+k)^{n-m}\tag {1}$$
The first few values are
$$f(1,m,a)=\frac{\Gamma (m-1)}{\Gamma (m)}\left(a^{1-m}-(a+1)^{1-m}\right)$$
$$f(2,m,a)=\frac{\Gamma (m-2)}{\Gamma (m)}\left(a^{2-m}-2 (a+1)^{2-m}+(a+2)^{2-m}\right)$$
$$f(3,m,a)=\frac{\Gamma (m-3)}{\Gamma (m)}\left(a^{3-m}-3 (a+1)^{3-m}+3 (a+2)^{3-m}-(a+3)^{3-m}\right)$$
$$f(4,m,a)=\frac{\Gamma (m-4)}{\Gamma (m)}\left(a^{4-m}-4 (a+1)^{4-m}+6 (a+2)^{4-m}-4 (a+3)^{4-m}+(a+4)^{4-m}\right)$$
Notice that (1) is valid for real $m$. If we want integer $m$ we need to take the limit. Here the $\Gamma$-function may have a pole singularity (infinity), which "competes" with the zero in the polynomial in $a$.
For instance
$$f(1,m\to 1,a)=\lim_{m\to 1} \, \frac{\left(a^{1-m}-(a+1)^{1-m}\right) \Gamma (m-1)}{\Gamma (m)}=\log (a+1)-\log (a)$$
For $n = 1, m=0 .. 4$ the limits are
$$\left( \begin{array}{cc} 0 & 1 \\ 1 & \log (a+1)-\log (a) \\ 2 & \frac{1}{a^2+a} \\ 3 & \frac{2 a+1}{2 a^2 (a+1)^2} \\ 4 & \frac{1}{3} \left(\frac{1}{a^3}-\frac{1}{(a+1)^3}\right) \\ \end{array} \right)$$
Or, a more complicated case,
$n=4, m = 0 .. 6$
$$\left( \begin{array}{cc} 0 & 1 \\ 1 & \frac{1}{6} \left(\log (a) a^3-4 (a+1)^3 \log (a+1)+6 (a+2)^3 \log (a+2)-4 (a+3)^3 \log (a+3)+(a+4)^3 \log (a+4)\right) \\ 2 & \frac{1}{2} \left(-a^2 \log (a)+4 (a+1)^2 \log (a+1)-6 (a+2)^2 \log (a+2)+4 (a+3)^2 \log (a+3)-(a+4)^2 \log (a+4)\right) \\ 3 & \frac{1}{2} (a \log (a)-4 (a+1) \log (a+1)+6 a \log (a+2)+12 \log (a+2)-4 a \log (a+3)\\&-12 \log (a+3)+a \log (a+4)+4 \log (a+4)) \\ 4 & \frac{1}{6} (-\log (a)+4 \log (a+1)-6 \log (a+2)+4 \log (a+3)-\log (a+4)) \\ 5 & \frac{1}{a^5+10 a^4+35 a^3+50 a^2+24 a} \\ 6 & \frac{1}{120} \left(-\frac{4}{(a+1)^2}+\frac{6}{(a+2)^2}-\frac{4}{(a+3)^2}+\frac{1}{(a+4)^2}+\frac{1}{a^2}\right) \\ \end{array} \right)$$
These examples show that our formula contains rather compact information which unfolds in these limits.
Derivation
The derivation starts with the formula
$$\int_0^{\infty } t^{m-1} e^{-s\; t} \, dt=s^{-m} \Gamma (m)\tag{2}$$
which generates the denominator of the original multiple integral
$$g(n,m,a)=\int _0^1...\int _0^1\frac{1}{s^m}dx_1 dx_2 ... dx_n\tag{3}$$
where
$$s = a + \sum _{i=1}^n x_i\tag{4}$$
Substituting (2) into (3) using (4) gives
$$g = \frac{1}{\Gamma(m)}\int_0^{\infty } t^{m-1} e^{-a\; t} (\prod _{i=1}^n \int _0^1 e^{-x_i\; t}\, dx_i )\, dt $$
Now the $x_i$ integrals can easily be done leaving us with
$$g = \frac{1}{\Gamma(m)}\int_0^{\infty } t^{m-1} e^{-a\; t} ((1-e^{- t})/t)^n \, dt $$
The final step, which is to show that $g$ is equal to $f$ given in the solution, is left as an exercise to the reader.
On
Use the integral representation $$ \frac{1}{(a+y)^m}=\frac{1}{\Gamma(m)}\int_0^\infty d\xi \xi^{m-1}e^{-\xi (a+y)} $$ to rewrite your integral as $$ I(m,n)=\int_0^1 \! \int_0^1 ...\int_0^1 \frac{dx_1dx_2...dx_n}{(a+x_1+x_2+...+x_n)^m}= \frac{1}{\Gamma(m)}\int_0^\infty d\xi \xi^{m-1}e^{-a \xi}\left[\int_0^1 dx e^{-\xi x}\right]^n $$ $$ =\frac{1}{\Gamma(m)}\int_0^\infty d\xi \xi^{m-1}e^{-a \xi}\left[\frac{1-e^{-\xi }}{\xi }\right]^n\ . $$ Expanding via the binomial theorem $$ I(m,n)=\frac{1}{\Gamma(m)}\sum_{k=0}^n {n\choose k}(-1)^{n-k}\int_0^\infty d\xi\ \xi^{m-1-n}e^{-\xi (a+n-k)} =\frac{1}{\Gamma(m)}\sum_{k=0}^n {n\choose k}(-1)^{n-k}\Gamma (m-n) (a-k+n)^{n-m}\ . $$ For certain combinations of $n$ and $m$, you'll need to evaluate the expression as a limit, for example for $n=2$ and $m=1$, you will need to evaluate $\lim_{m\to 1} I(m,2)$, which coincides with the value found above.
Let $$ f_{n,m}(a)=\int_0^1\cdots\int_0^1\frac{\prod_n dx_i}{(a+\sum_n x_i)^m} $$ Now, it is easy enough to confirm that $$ \frac{df_{n,m}}{da}=-mf_{n,m+1}(a) $$ It is also easy enough to see that $$ f_{n+1,m}(a)=\int_0^1 f_{n,m}(x+a)\ dx \tag{1} $$ which, on taking the derivative, gives $$ f_{n+1,m}'(a) = f_{n,m}(a+1)-f_{n,m}(a) $$ Bringing these together gives $$ -mf_{n+1,m+1}(a) = f_{n,m}(a+1)-f_{n,m}(a) = \Delta f_{n,m}(a) $$ where $\Delta$ refers to the forward difference, or $$ f_{n+1,m+1}(a)=-\frac{\Delta f_{n,m}(a)}{m} $$ Iterating down, we can see that, for non-integer $m$, and also for integer $m\geq n$, $$ f_{n,m}(a)=(-1)^{n-1}\frac{\Gamma(m-n+1)}{\Gamma(m)}\Delta^{n-1} f_{1,m-n+1}(a) $$ Now, we can calculate for $k\neq1$, $$ f_{1,k}(a)=\int_0^1 \frac{dx}{(a+x)^k} = \frac1{k-1}\left(\frac1{a^{k-1}}-\frac1{(a+1)^{k-1}}\right) = \frac{\Delta(a^{1-k})}{1-k} $$ and $$ f_{1,1}(a)=\log(a+1)-\log(a) $$
To deal with integer $m\leq n$, we make use of equation (1), and it can be confirmed by induction that $$ f_{n,1}(a) = \frac{\Delta^n (a^{n-1}\log(a))}{\Gamma(n)} \tag{2} $$ and we can then write $$ f_{n,m}(a)=(-1)^{m-1}\frac{\Delta^{m-1}f_{n-m+1,1}(a)}{\Gamma(m)} $$ and so, the general form of the solution is $$ f_{n,m}(a)=\begin{cases} \frac{(-1)^{m-1}}{\Gamma(n-m+1)\Gamma(m)}\Delta^{n} (a^{n-m}\log(a)) &n\geq m\in\mathbb{Z}^+\\ (-1)^n\frac{\Gamma(m-n)}{\Gamma(m)}\Delta^n (a^{n-m}) & \text{otherwise} \end{cases} $$ where for integer $m\leq0$, the ratio of gamma functions should be evaluated as one would a ratio of negative factorials.
The Induction (how to get/confirm equation (2)):
The key to this equation is that, when you apply equation (1), the polynomial terms get cancelled out by the repeated Forward Difference operator application.
I hypothesised the form of the equation by examining the first few terms (first one as seen above, second and third both provided in question):
$$\begin{align} f_{1,1}(a)&=\log(a+1)-\log(a)=\Delta(\log(a))\\ f_{2,1}(a)&=(a+2)\log(a+2)-2(a+1)\log(a+1)+a\log(a) = \Delta^2(a\log(a))\\ f_{3,1}(a)&=\frac{(a\!+\!3)^2\log(a\!+\!3)-3(a\!+\!2)^2\log(a\!+\!2)+3(a\!+\!1)^2\log(a\!+\!1)-a^2\log(a)}{2}\\ &=\frac{\Delta^3(a^2\log(a))}{2} \end{align}$$ Each power of $a$ in front of the $\log$ terms is one less than $n$, the $\Delta$ iteration matches $n$, and the denominator appears to be following a factorial pattern (1,1,2) - checking $n=4$ (see Dr. Wolfgang Hintze's answer for its expression) can help to confirm the factorial pattern, as the denominator is 6.
So we hypothesise $$ f_{n,1}(a)=\frac{\Delta^n(a^{n-1}\log(a))}{(n-1)!}=\frac{\Delta^n(a^{n-1}\log(a))}{\Gamma(n)} $$ where it has been expressed with $\Gamma$ in order to match the form used for other cases.
Now for the induction to confirm. We can see that it's right for $n=1$ easily, so I'll focus on the induction step. Assume $f_{n,1}(a)$ fits that form - can we confirm that $f_{n+1,1}(a)$ also fits that form?
Using equation (1), and noting that the Forward Difference operator commutes with integration, $$\begin{align} f_{n+1,1}(a)&=\int_0^1 f_{n,1}(x+a)\ dx\\ &=\frac{\Delta^n}{\Gamma(n)}\int_0^1 (x+a)^{n-1}\log(x+a)\ dx \end{align}$$ We apply integration by parts with $u=\log(x+a)$ and $dv=(x+a)^{n-1}dx$ to get $$\begin{align} \int_0^1 (x+a)^{n-1}\log(x+a)\ d&x=\left[\frac{(x+a)^n}n\log(x+a)\right]_0^1-\int_0^1 \frac{(x+a)^{n-1}}ndx\\ &=\frac{\Delta(a^n\log(a))}n - \left[\frac{(x+a)^n}{n^2}\right]_0^1\\ &=\frac{\Delta(a^n\log(a))}n - \frac{\Delta(a^n)}{n^2} \end{align}$$ And so we have $$\begin{align} f_{n+1,1}(a)&=\frac{\Delta^n}{\Gamma(n)}\left(\frac{\Delta(a^n\log(a))}n - \frac{\Delta(a^n)}{n^2}\right)\\ &=\frac{1}{n\Gamma(n)}\left(\Delta^{n+1}(a^n\log(a)) - \frac{\Delta^{n+1}(a^n)}{n}\right) \end{align}$$ Now, it can be easily confirmed that $\Delta^n(a^m)=0$ if $n>m\geq0$ and $m$ is an integer (because $\Delta(a^m)$ is a polynomial of order $m-1$). Also, $\Gamma(n+1)=n\Gamma(n)$. So we have $$ f_{n+1,1}(a)=\frac{\Delta^{n+1}(a^n\log(a))}{\Gamma(n+1)} $$ which is what we needed to show. By induction, this proves our formula.