Where am I wrong in this prove of concavity?

Question

I have following function $$f(x_0,x_1)=-x_1Ei(-a-\frac{bx_1}{x_0})\exp(a+\frac{bx_1}{x_0})$$. Where $a,b$ are greater than zero, $x_0,x_1$ can have positive values and $Ei(x)$ is the exponential integral. Now its perspective function is $$f_1(x_0)=-Ei(-a-\frac{b}{x_0})\exp(a+\frac{b}{x_0})$$ If I do following transformation $\frac{1}{t}=a+\frac{b}{x_0}$ then I have $$f_2(t)=-Ei(-\frac{1}{t})\exp(\frac{1}{t})$$ Now $f_2(t)$ is concave with respect to $t$. Based on this I assume that $f_1(x_0)$ is concave with respect to $x_0$. Since $f_1(x_0)$ is the perspective function of $f(x_0,x_1)$ therefore I conclude that $f(x_0,x_1)$ is jointly concave with respect to $(x_0,x_1)$. But when I fix $x_0$ and try to prove the concavity of $f(x_0,x_1)$ with respect to $x_1$ through double derivative test. Then I am unable to prove this. Is the function $f(x_0,x_1)$ not jointly concave because if it were then it should be concave with respect to $x_1$ also. Any help in this regard will be much appreciated. Thanks in advance.

Answer 1

\begin{align}f_1(x_0)&=-Ei(-a-\frac{b}{x_0})\exp(a+\frac{b}{x_0})\\ &=-Ei\left(\color{red}- \frac1t\right) \exp\left(\frac1t\right)\end{align}