Series expansion around natural logarithm

Question

I am working on an integral using the Laplace-method, and I have to do a series expansion of the following

$$ \phi(x,t) = x \ln(t) - t, $$

according to the solution the answer is

$$ \phi(x,t) = x \ln(x) - x - \frac{1}{2x}(t-x)^2 + \dots. $$

The only comment they mention in the text is that this is an expansion, but I am not sure as to how I can arrive at this result. Could anyone help me?

Answer 1

Consider $$\ln\frac tx=\ln\left(1+\frac{t-x}x\right)=\sum_{i=1}^\infty\frac{(-1)^{i-1}}i\left(\frac{t-x}x\right)^i\\=\frac{t-x}x-\frac12\left(\frac{t-x}x\right)^2+\frac13\left(\frac{t-x}x\right)^3-\frac14\left(\frac{t-x}x\right)^4\dots$$

Then $$x\ln t-t=x(\ln x+\ln\frac tx)-t=x\ln x+x\sum_{i=1}^\infty\frac{(-1)^{i-1}}{ix^i}(t-x)^i-t\\= x\ln x-t+(t-x)-\frac1{2x}\left(t-x\right)^2+\frac1{3x^2}\left(t-x\right)^3-\frac1{4x^3}\left(t-x\right)^4\dots$$

Answer 2

Built around $t=x$, Taylor expansion of $\log(t)$ is simply $$\log(t)=\log (x)+\frac{t-x}{x}-\frac{(t-x)^2}{2 x^2}+\frac{(t-x)^3}{3 x^3}+O\left((t-x)^4\right)$$

I let you the remaining of the work.