my task is to prove that $\int_{0}^{\infty} \frac{t^a}{sinh(t)} \frac{x}{x^2+t^2} dt= x^a \int_{0}^{\infty} \frac{t^a}{sinh(xt)} \frac{1}{1+t^2} dt$
so starting with the integral on the right hand side and by substituting u=xt one gets $x^a \int_0^{\infty} \frac{\frac{u^a}{x^a}}{sinh(u)} \frac{1}{1+ (\frac{u}{x})^2} \frac{1}{x} du$ which in the end is equal to $\int_{0}^{\infty} \frac{u^a}{sinh(u)} \frac{x}{x^2+u^2} du$. So my question is: Is this a legitimate proof since we can rename u=t?
I'm going to assume that $x>0$, so \begin{align} x^a \int_{0}^{\infty} \frac{t^a}{\sinh(xt)} \frac{1}{1+t^2} dt&=x^a \lim_{b\to+\infty}\int_{0}^{b} \frac{t^a}{\sinh(xt)} \frac{1}{1+t^2} dt\qquad(\text{doing $u=xt$})\\ &=x^a \lim_{b\to+\infty}\int_{0}^{bx} \frac{\frac{u^a}{x^a}}{\sinh(u)} \frac{1}{1+ (\frac{u}{x})^2} \frac{1}{x} du\\ &=\lim_{b\to+\infty}\int_{0}^{bx} \frac{u^a}{\sinh(u)} \frac{x}{x^2+ u^2}du\\ &=\lim_{b\to+\infty}\int_{0}^{bx} \frac{t^a}{\sinh(t)} \frac{x}{x^2+ t^2}dt\\ &=\int_{0}^{+\infty} \frac{t^a}{\sinh(t)} \frac{x}{x^2+ t^2}dt \end{align}
In case $x<0$, then we arrive at the following: \begin{align} x^a \int_{0}^{\infty} \frac{t^a}{\sinh(xt)} \frac{1}{1+t^2} dt&=\lim_{b\to+\infty}\int_{0}^{bx} \frac{u^a}{\sinh(u)} \frac{x}{x^2+ u^2}du\\ &=\int_{0}^{-\infty} \frac{u^a}{\sinh(u)} \frac{x}{x^2+ u^2}du\qquad(\text{doing $t=-u$})\\ &=\int_{0}^{+\infty} \frac{(-1)^a t^a}{-\sinh(t)} \frac{x}{x^2+ t^2}(-dt)\\ &=(-1)^a\int_{0}^{+\infty} \frac{t^a}{\sinh(t)} \frac{x}{x^2+ t^2}dt \end{align}
