Closed-form solution for an integral with integer power rational function I need a closed-form solution for the following integral:
$$\int_{0}^{\infty}\frac{dt}{(t+a)^{m} (t +b)^{n}};\,\,a,b>0;\,\,m,n\geq1\,\text{are integers}$$
If $m$ and $n$ are non-integers, I can find solutions in books. 
I tried with partial fraction decomposition. But it does not help as I get these integrals at some point: $\int_{0}^{\infty}\frac{dt}{(t+a)}$ and $\int_{0}^{\infty}\frac{dt}{(t +b)}$ which may not have solutions too. 
Is there any  other solution? 
 A: Everything can be computed from the $m=n=1$ case by differentiating with respect to $a$ and $b$:
$$ \frac{1}{(x+a)^m} = \frac{(-1)^{m-1}}{m!} \frac{\partial^{m-1}}{\partial a^{m-1}} \frac{1}{x+a}, $$
so if we write $I_{m,n}(a,b)$ for your integral,
$$ I_{m,n}(a,b) = \frac{(-1)^{m-1}}{m!n!} \frac{\partial^{m+n-2}}{\partial a^{m-1}\partial b^{n-1}} I_{1,1}(a,b). $$
$I_{1,1}$ is easy to compute: we have
$$ I_{1,1}(a,b) = \frac{\log{a}-\log{b}}{a-b}, $$
writing it symmetrically. Now we can apply Leibniz's formula
$$ D^n(fg) = \sum_{k=0}^n \binom{n}{k} (D^kf)(D^{n-k}g) $$
to work out the derivatives. In particular, 
$$ \frac{(-1)^{m-1}}{m!} \frac{\partial^{m-1}}{\partial a^{m-1}} \frac{\log{a}-\log{b}}{a-b} \\
= \frac{\log{a}-\log{b}}{(a-b)^m} + \frac{(-1)^{m-1}}{m!} \sum_{k=1}^{m-1} \binom{m-1}{k} \frac{(-1)^{m-1-k} (m-k)!}{(a-b)^{m-k}} \frac{ (-1)^{k-1}(k-1)! }{a^k} \\
= \frac{\log{a}-\log{b}}{(a-b)^m} - \sum_{k=1}^{m-1} \frac{(m-1)!}{k!(m-k-1)!}\frac{ (m-k)!(k-1)!}{m!} \frac{1}{a^k(a-b)^{m-k}} \\
= \frac{\log{a}-\log{b}}{(a-b)^m} - \sum_{k=1}^{m-1} \left(\frac{1}{k}-\frac{1}{m} \right)\frac{1}{a^k(a-b)^{m-k}}. $$
One can now differentiate with respect to $b$ to get a couple of finite series and double series for $I_{m,n}(a,b)$.
Alternatively, one can combine this expression, which is effectively $I_{m,1}(a,b)$, with a recurrence relation from integration by parts, namely
$$ \frac{1}{a^mb^n} + m I_{m+1,n}(a,b) + n I_{m,n+1}(a,b) = 0, $$
which gives the answer in terms of only single series. Sadly we do seem to have lost symmetry between $a$ and $b$ in this process.
A: In the partial fraction decomposition, the coefficients of $1/(t+a)$ and $1/(t+b)$ should cancel.  Note that if $a, b > 0$
$$ \int_0^\infty \left(\frac{1}{t+a} - \frac{1}{t+b}\right)\; dt 
= \lim_{R \to \infty} \ln(R+a) - \ln(R+b) - \ln(a) + \ln(b) = \ln(b/a)$$
A: In what follows we suppose that $a\ne b$. Let
$$F(X)=
\frac{1}{(X+a)^m(X+b)^n}$$
and $I(n,m)=\int_0^\infty F(t)dt$. Now, for $x$ near $0$ we have
$$\eqalign{F(x-a)&=\frac{1}{x^m(x+b-a)^n}=\frac{1}{(b-a)^nx^m}\left(1+\frac{x}{b-a}\right)^{-n}\cr
&=\frac{1}{(b-a)^nx^m}\left(\sum_{k=0}^{m-1}\binom{-n}{k}\frac{x^k}{(b-a)^{k}}+\mathcal{O}(x^m)\right)\cr
&=\sum_{k=0}^{m-1}\binom{n+k-1}{k}\frac{(-1)^k}{x^{m-k}(b-a)^{n+k}}+\mathcal{O}(1)}$$
So, the polar part of $F$ corresponding to the pole $-a$ is
$$Q_{-a}(X)=\sum_{k=0}^{m-1}\binom{n+k-1}{k}\frac{(-1)^k}{(X+a)^{m-k}(b-a)^{n+k}}$$
Exchnageing $(a,m)$ and $(b,n)$ we get the polar part of $F$ corresponding to the pole $-b$
$$Q_{-b}(X)=\sum_{k=0}^{n-1}\binom{m+k-1}{k}\frac{(-1)^k}{(X+b)^{n-k}(a-b)^{m+k}}$$
This yields the partial fraction decomposition:
$$
F(X)=\binom{n+m-2}{m-1}\frac{(-1)^{m-1}}{(b-a)^{n+m-1}}\left(\frac{1}{X+a}-\frac{1}{X+b}\right)
+\sum_{k=0}^{m-2}\binom{n+k-1}{k}\frac{(-1)^k}{(X+a)^{m-k}(b-a)^{n+k}}
+\sum_{k=0}^{n-2}\binom{m+k-1}{k}\frac{(-1)^k}{(X+b)^{n-k}(a-b)^{m+k}}
$$
So,
$$
I(n,m)=\binom{n+m-2}{m-1}\frac{(-1)^{m-1}}{(b-a)^{n+m-1}}\ln\left(\frac{b}{a}\right)
+\sum_{k=0}^{m-2}\binom{n+k-1}{k}\frac{(-1)^k}{(m-k-1)a^{m-k-1}(b-a)^{n+k}}
+\sum_{k=0}^{n-2}\binom{m+k-1}{k}\frac{(-1)^k}{(n-k-1)b^{n-k-1}(a-b)^{m+k}}
$$
