Closed-form solution for an integral with integer power rational function

Question

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?

Answer 1

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, $$ I(n,m)=\binom{n+m-2}{m-1}\frac{(-1)^{m-1}}{(b-a)^{n+m-1}}\ln\left(\frac{b}{a}\right) +\sum_{k=0}^{m-2}\binom{n+k-1}{k}\frac{(-1)^k}{(m-k-1)a^{m-k-1}(b-a)^{n+k}} +\sum_{k=0}^{n-2}\binom{m+k-1}{k}\frac{(-1)^k}{(n-k-1)b^{n-k-1}(a-b)^{m+k}} $$

Answer 2

In the partial fraction decomposition, the coefficients of $1/(t+a)$ and $1/(t+b)$ should cancel. Note that if $a, b > 0$

$$ \int_0^\infty \left(\frac{1}{t+a} - \frac{1}{t+b}\right)\; dt = \lim_{R \to \infty} \ln(R+a) - \ln(R+b) - \ln(a) + \ln(b) = \ln(b/a)$$

Answer 3

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.