I would like to find an equivalent (first term of an asymptotic expansion) of the following function in $x=0$ and $x=1$:
$$f(x)=\int_0^{+\infty} \frac{1}{t^x\sqrt{1+t^2}}\ dt$$
My professor gave a highly unmotivated solution which doesn't invoke the usual tools to find asympto tic expansions of integrals depending on some parameter.
I would be interested in an elementary, but motivated solution.
Herein we present a way forward, which relies on elementary analysis only, to derive the first three terms of the "small $x$" expansion of the function of interest. To that end, we now proceed.
For $x>0$, we have
$$\begin{align} f(x)&=\int_0^\infty \frac{e^{-x\log(t)}}{\sqrt{t^2+1}}\,dt\\\\ &=\int_0^1 \frac{e^{-x\log(t)}}{\sqrt{t^2+1}}\,dt+\int_1^\infty \frac{e^{-x\log(t)}}{\sqrt{t^2+1}}\,dt\\\\ &=\int_0^1 \frac{e^{-x\log(t)}}{\sqrt{t^2+1}}\,dt+\int_0^1 \frac{e^{x\log(t)}}{t\sqrt{t^2+1}}\,dt\tag1 \end{align}$$
The first term on the right-hand side of $(1)$ is well behaved near $x=0^+$ with
$$\begin{align} \int_0^1 \frac{e^{-x\log(t)}}{\sqrt{t^2+1}}\,dt&=\text{arsinh}(1)-x\int_0^1 \frac{\log(t)}{\sqrt{1+t^2}}\,dt+O(x^2)\\\\ &=\text{arsinh}(1)-x\int_0^{\pi/4} \log(\tan(\phi))\,\sec(\phi)\,d\phi+O(x^2)\tag2 \end{align}$$
The second term on the right-hand side of $(1)$ can be written as
$$\begin{align} \int_0^1 \frac{e^{x\log(t)}}{t\sqrt{t^2+1}}\,dt&=\int_0^1 \frac{e^{x\log(t)}}{t}\,dt-\int_0^1 \frac{e^{x\log(t)}}{t}\left(1-\left(1+t^2\right)^{-1/2}\right)\,dt\\\\ &=\frac1x -\int_0^1 \frac{1-\left(1+t^2\right)^{-1/2}}{t}\,dt-x\int_0^1 \frac{(1-\left(1+t^2\right)^{-1/2})\,\log(t)}{t}\,dt+O(x^2)\\\\ &=\frac1x -\text{arsinh}(1)+\log(2)-x\int_0^{\pi/4}\log(\tan(\phi))\,\sec(\phi)\frac{1-\cos(\phi)}{\sin(\phi)}\,d\phi+O(x^2)\tag3 \end{align}$$
Substituting $(2)$ and $(3)$ into $(1)$ reveals
$$\begin{align} f(x)&=\frac1x +\log(2) -x\int_0^{\pi/4}\log(\tan(\phi))\,\sec(\phi)\left(1+\frac{1-\cos(\phi)}{\sin(\phi)}\right)\,d\phi+O(x^2)\\\\ &=\frac1x +\log(2) +\frac1{12}\left(\pi^2+6\log^2(2)\right)\,x+O(x^2) \end{align}$$
This looks like a rather tedious calculation, so resorting to computer algebra we find (for $0 < x < 1$):
$f(x) = \frac{\Gamma \left(\frac{1}{2}-\frac{x}{2}\right) \Gamma \left(\frac{x}{2}\right)}{2 \sqrt{\pi }}$.
Notice that it is symmetric with respect to $x = 1/2$.
The first three terms in the expansion around $x=0$ are:
$f(x) \approx \frac{1}{x}+\log (2)+\frac{1}{12} x \left(\pi ^2+6 \log ^2(2)\right)+\frac{1}{12} x^2 \left(3 \zeta (3)+2 \log ^3(2)+\pi ^2 \log (2)\right)+O\left(x^3\right)$
$f$ is defined in $(0,1)$, and it's symmetric with respect to $x={1\over2}$ since \begin{align} f(x)&=\int_0^{1} \frac{1}{t^x\sqrt{1+t^2}}dt+\int_1^{+\infty} \frac{1}{t^x\sqrt{1+t^2}}dt\\ &=\int_0^{1} \frac{1}{t^x\sqrt{1+t^2}}dt+\int_0^{1} \frac{1}{s^{1-x}\sqrt{1+s^2}}ds\\ \end{align} thus we can study on only one side, here we choose $x\to0^+$, then we have \begin{align} f(x)&=\int_0^{1} \frac{1}{s^{1-x}\sqrt{1+s^2}}ds+\operatorname{arsinh}(1)+o(1)\\ &=[{s^x\over x}\frac{1}{\sqrt{1+s^2}}]_0^1+{1\over x}\int_0^{1}\frac{s^{x+1}}{(1+s^2)^{3/2}}ds+\ln(1+\sqrt{2})+o(1)\\ &={1\over \sqrt{2}x}+{1\over x}(1-{1\over \sqrt{2}})+o({1\over x})\\ &={1\over x}+o({1\over x}) \end{align}
