Does this integral have simple result expression?

Question

I am evaluating the integral

$$\int_0^\infty\frac{1}{(1+x)^a(1+bx)^2}\,\mathrm{d}x,$$

where $a>0$ and $b>1$.

By expressing $(1+x)^{-a}$ and $(1+bx)^2$ in terms of Meijer-G function, I can obtain the result for the above integral, but the result contains the Meijer-G function.

I wonder if the above integral can have a simple expression result, i.e., without containing any special functions.

Thanks!

Answer 1

Except if $a$ is an integer, you will face hypergeometric functions such as $$\frac{1+\frac{a \, _2F_1(1,1;2-a;b)}{1-a}-a\pi (1-b)^{-a} \left(\frac{1}{b}\right)^{1-a} \csc (\pi a)}{b-1}$$

If $a$ is an integer, the problem is quite different. Let $$I_n=\int_0^\infty\frac{1}{(1+x)^n(1+bx)^2}\,dx$$ $$ \,I_n=\frac{(b-1)P_{n-1}(b)-n b^{n-1}\log(b)}{(b-1)^n }$$ where $P_k$ is a polynomial of degree $k$ in $b$.

In these polynomials, the constant term is $\frac{(-1)^n}{n-1}$ and the coefficient of $b^{n-1}$ is $1$.

What are these polynomials ?

Answer 2

After Claude Leibovici's answer, let me calculate $$J_n(B) = \int_0^\infty \frac{1}{(1+x)^n (B+x)^2} dx$$ when $n$ is an integer $\ge 2$. (Note that $I_n$ and $J_n(B)$ are different; we have $$I_n = \int_0^\infty \frac{1}{(1+x)^n (1+bx)^2} dx = \frac{1}{b^2}\int_0^\infty \frac{1}{(1+x)^n (b^{-1}+x)^2} dx = b^{-2}J_n(b^{-1})$$ and $J_n$ is more convenient to deal with.)

One can assert $$ \frac{1}{(1+x)^n(B+x)^2} = \frac{q_n(x)}{(1+x)^n} + \frac{1}{(B+x)^2}(\alpha + \beta (B+x))$$ where $q_n$ is some polynomial of degree $\le n-1$ and $\alpha,\beta$ are constant. Multiply $(1+x)^n (B+x)^2$ to the both side, put $x = -B$ to have $\alpha = (1-B)^{-n}$, differentiate with respect to $x$ and put $x= - B$ to have $\beta = -n (1-B)^{-n-1}$; thus we have

$$ \frac{(1-B)^{n+1}}{(1+x)^n(B+x)^2} = \frac{\tilde{q}_n(x)}{(1+x)^n} + \frac{1}{(B+x)^2}(1-B -n (B+x))$$ where $\tilde{q}$ is some polynomial of degree $\le n-1$.

Now

\begin{align*} (B+x)^2\tilde{q}_n(x)&= (1-B)^{n+1} - \left(1-B -n (B+x))\right)(1+x)^n\\ & =(1-B)^{n+1} +\left(nx -1+B +nB)\right)(1+x)^n\\ x^2 \tilde{q}_n(x-B) & =(1-B)^{n+1} +\left(nx -1+B)\right)(x+1 - B)^n \end{align*} one can apply the binomial theorem and do bunch of calculations to get \begin{align*} \tilde{q}_n(x) &= \sum_{k=0}^{n-1} (k+1) \binom{n+1}{k+2} (1-B)^{n-k-1} (x+B)^k\\ & = \sum_{j=0}^{n-1}a_{n, j}(1-B)^{n-j-1}(1+x)^j \end{align*} where $$a_{n, j}= \sum_{k=0}^{n-j-1} (k+1) \binom{n+1}{k+j+2} \binom{k+j+1}{j}(-1)^k.$$ One can check that $a_{n, j}=j+1$.

Thus, \begin{align*} (1-B)^{n+1} J_n(B)&= (1-B)\int_0^\infty\frac{1}{(x+B)^2} dx - n \int_0^\infty\frac{1}{x+B}dx+\\ &\phantom{=+}+\sum_{j=0}^{n-2}a_{n, j}(1-B)^{n-j-1}\int_0^\infty \frac{1}{(1+x)^{n-j}}dx + n \int_{0}^\infty \frac{1}{x+1}dx\\ & = \frac{1-B}{B} + n \log B+(1-B)\sum_{j=0}^{n-2}(1-B)^{n-2-j}\frac{j+1}{n-1-j} \\ J_n(B)&= \frac{1-B +nB \log B}{B(1-B)^{n+1}} + \frac{1}{(1-B)^n}\sum_{j=0}^{n-2}(1-B)^{n-2-j}\frac{j+1}{n-1-j} \end{align*}

Note that the coefficient in the last summation term $\frac{j+1}{n-1-j} $ is symmetric, in the sense of $\frac{j+1}{n-1-j} \cdot \frac{(n-2-j)+1}{n-1-(n-2-j)} =1$.