$0<b<a$
$$I=\int_{-\infty}^{\infty}xe^{-\pi{x^2}\left(a+x\over b+x\right)^2}dx=b-a.$$
Applying integration by parts here is doesn't work.
$u=x$ then $du=dx$
$dv=e^{-\pi{x^2}\left(a+x\over b+x\right)}dx$ Then
$v=\int_{-\infty}^{-\infty}e^{-\pi{x^2}\left(a+x\over b+x\right)}dx=1$
$$I=x-\int_{-\infty}^{-\infty}dx$$
It doesn't make any sense here.
Can anyone provide a prove of this integral?
Differentiating the equation $$\int_{-\infty}^{+\infty}f\left(x-\frac{a}{x-b}\right)dx=\int_{-\infty}^{+\infty} f(x)dx,\qquad a>0$$ with respect to $a$ and $b$ one obtains $$\int_{-\infty}^{+\infty}\frac{1}{x-b}f'\left(x-\frac{a}{x-b}\right)dx=0\tag{1} $$ $$\int_{-\infty}^{+\infty}\frac{1}{(x-b)^2}f'\left(x-\frac{a}{x-b}\right)dx=0\tag{2} $$ Also $$\int_{-\infty}^{+\infty}f'\left(x-\frac{a}{x-b}\right)dx=0\tag{3} $$ Now consider $$I(a,b)=\int_{-\infty}^{+\infty}xf\left(x-\frac{a}{x-b}\right)dx. $$ \begin{align} \frac{\partial I}{\partial a}&=-\int_{-\infty}^{+\infty}\frac{x}{x-b}f'\left(x-\frac{a}{x-b}\right)dx\\ &=-\int_{-\infty}^{+\infty}\left(1+\frac{b}{x-b}\right)f'\left(x-\frac{a}{x-b}\right)dx=0. \quad \text{(due to (1) and (3))} \end{align} \begin{align} \frac{\partial I}{\partial b}&=-\int_{-\infty}^{+\infty}\frac{x}{(x-b)^2}f'\left(x-\frac{a}{x-b}\right)dx\\ &=-\int_{-\infty}^{+\infty}\left(\frac{1}{x-b}+\frac{b}{(x-b)^2}\right)f'\left(x-\frac{a}{x-b}\right)dx=0. \quad \text{(due to (1) and (2))} \end{align} Thus $$I(a,b)=\int_{-\infty}^{+\infty} xf(x)dx$$ Now for $\alpha>\beta$ one has \begin{align} \int_{-\infty}^{+\infty}xf\left(x\frac{\alpha +x}{\beta +x}\right){d}x&=\int_{-\infty}^{+\infty}xf\left(x-\frac{(\alpha-\beta)\beta}{x+\beta}+\alpha-\beta\right)dx\\ &=\int_{-\infty}^{+\infty}(x+\beta-\alpha)f\left(x-\frac{(\alpha-\beta)\beta}{x+2\beta-\alpha}\right)dx\\ &=I((\alpha-\beta)\beta,\alpha-2\beta)+(\beta-\alpha)\cdot \int_{-\infty}^{+\infty} f(x)dx \end{align} Finally
$$ \int_{-\infty}^{+\infty}xf\left(x\frac{\alpha +x}{\beta +x}\right)dx=\int_{-\infty}^{+\infty} xf(x)dx+(\beta-\alpha)\cdot \int_{-\infty}^{+\infty} f(x)dx,\quad \alpha>\beta. $$
Here's a general approach to treat integrals of the form $$ \int_{-\infty}^\infty x^{s-1}f(u)\text d x $$ Where $s$ is a nonzero integer and for a set of positive real numbers $\{a_k\}$, a set of increasing real numbers $\{b_k\}$, $$ u(x)=x-\sum_{k=1}^{n-1}\frac{a_k}{x-b_k}. $$ Note that $$ u'(x)=1+\sum_{k=1}^{n-1}\frac{a_k}{(x-b_k)^2}>0 $$ so $u(x)$ increases intervalwise-monotonically and hence has inverse in each interval $(-\infty,b_1),(b_1,b_2),\cdots, (b_{n-1},\infty)$, denoted by $x_1(u),x_2(u),\cdots, x_{n}(u)$ respectively. With the help of the graph, one easily verifies $u(x)$ maps each interval to $\mathbb R$. Split the real line into the intervals and take the inverse respectively, $$ \begin{align} &\int_{-\infty}^\infty x^{s-1}f(u(x))\text d x \\ =&\int_{-\infty}^{b_1}x^{s-1}_1f(u(x_1))\text d x_1+\int_{b_1}^{b_2}x^{s-1}_2f(u(x_2))\text d x_2+\cdots+\int_{b_{n-1}}^{\infty}x^{s-1}_{n}f(u(x_{n}))\text d x_{n} \\ =&\int_{-\infty}^\infty x^{s-1}f(u)\frac{\text d x_1}{\text d u}\text d u+\int_{-\infty}^\infty x^{s-1}f(u)\frac{\text d x_2}{\text d u}\text d u+\cdots+\int_{-\infty}^\infty x^{s-1}f(u)\frac{\text d x_{n}}{\text d u}\text d u \\ =&\frac1s\int_{-\infty}^\infty f(u)\frac{\text d }{\text d u}\Big(x_1^s+\cdots+x_{n}^s\Big)\text d u \\ =&\frac1s\int_{-\infty}^\infty f(u)\frac{\text d }{\text d u}p_s(\{x_k(u)\})\text d u \end{align} $$ Here $p_s(\{x_k(u)\})$ denotes the $s$ th power sum.
Consider the equation $$ u(x)=x-\sum_{k=1}^{n-1}\frac{a_k}{x-b_k}=u, $$ simple manipulation yields the algebraic equation $$ (x-b_1)\cdots(x-b_{n-1})(x-u)-\sum_{k=1}^na_k\prod_{\substack{j=1\\j\ne k}}^{n-1}(x-b_k)=0 $$ It's easy to see that the $n$ roots of the equation are exactly the inverse functions $x_1(u),x_2(u),\cdots, x_{n}(u)$.
The elementary symmetric polynomials of the roots are the coefficients of the equation (signs considered), so are polynomials of $u$. By the known property of elementary symmetric polynomials, every symmetric polynomial, of course including the power sum, can be expressed in polynomials of them. We can conclude that $p_s(\{x_k(u)\})$ is a polynomial of $u$ and the desired integral is reduced into a linear combination of $$ \int_{-\infty}^\infty u^{s'-1}f(u(x))\text d u, \quad s'=1,\cdots,s $$ so the problem is greatly simplified.
Corresponding integrals to $s=1,2,3$ are $$ \int_{-\infty}^\infty f(u)\text d x=\int_{-\infty}^\infty f(u)\text d u\\ \int_{-\infty}^\infty xf(u)\text d x=\int_{-\infty}^\infty uf(u)\text d u\\ \int_{-\infty}^\infty x^2f(u)\text d x=\int_{-\infty}^\infty u^2f(u)\text d u+\left(\sum_ka_k\right)\int_{-\infty}^\infty f(u)\text d u $$ Values of higher orders may be more tedious to work out. One may notice that the first one is the Glasser's Master Theorem.
Using the formula on the second line, with the same derivation as Nemo's, the identity holds $$ \int_{-\infty}^{\infty}xf\left(x\frac{a +x}{b +x}\right)\text dx=\int_{-\infty}^{\infty} xf(x)\text dx+(b-a)\int_{-\infty}^{\infty} f(x)\text dx,\quad |b|\le|a| $$ and the final result follows from the Gaussian integral.
