What is $\lim_{x\to 0^+} \int_0^\infty \frac{e^{-xt}}{\sqrt{t^2 +t}} dt$? What is $\lim_{x\to 0^+} \int_0^\infty \frac{e^{-xt}}{\sqrt{t^2 +t}} dt$ ?
I would like to say that it is $+\infty$. Though I don't know how to properly prove it.
I showed that $x\to\int_0^\infty \frac{e^{-xt}}{\sqrt{t^2 +t}} dt$ is derivable on $\mathbb R^+_*$ with the derivate : $x \to -\int_0^\infty e^{-xt} \sqrt{1-\frac{1}{t+1}}dt$
I don't know if there is a nice substitution.
EDIT : Also is it possible to find an equivalent when $x\to 0$ ?
 A: You can do the following:
$$\int_0^\infty\frac{e^{-xt}}{\sqrt{t^2+t}}\mathrm{d}t = \int_0^{\frac{1}{2x}}\frac{e^{-xt}}{\sqrt{t^2+t}}\mathrm{d}t + \int_{\frac{1}{2x}}^\infty\frac{e^{-xt}}{\sqrt{t^2+t}}\mathrm{d}t \geq \int_0^{\frac{1}{2x}}\frac{\frac{1}{2}}{\sqrt{t^2+t}}\mathrm{d}t+0.$$
Now use your favorite table of standard integrals/wolframalpha to solve this integral, and you obtain $\int\frac{1}{\sqrt{t^2+t}}\mathrm{d}t = 2\mathrm{arcsinh}(t)+c$, which diverges as $\frac{1}{2x}\to\infty$, i.e. as $x\to0^+$.
A: It is not difficult to show that the leading asymptotic term $\sim -\ln x$. Indeed, the integral converges at $t\to 0$, and at $t\to\infty\,\, \frac{1}{\sqrt{t^2 +t}}\to\frac{1}{t}$, what leads to logarithmic divergence. We can also show it explicitly.
$$I(x)=\int_0^\infty\frac{e^{-xt}}{\sqrt t\sqrt{t+1}}dt=2\int_0^\infty\frac{e^{-xs^2}}{\sqrt{1+s^2}}ds$$
Making the substitution $s=\sinh t$
$$=2\int_0^\infty e^{-x\sinh^2t}dt=2e^\frac{x}{2}\int_0^\infty e^{-\frac{x}{2}\cosh2t}dt=e^\frac{x}{2}K_0\Big(\frac{x}{2}\Big)$$
Now we can use the known asymptotic, or just get it. Using the substitution $z=e^t$,
$$I(x)=e^\frac{x}{2}\int_1^\infty e^{-\frac{x}{4}\big(z+\frac{1}{z}\big)}\frac{dz}{z}$$
At $x\to0$ we are allowed to make a decomposition
$$=e^\frac{x}{2}\int_1^\infty e^{-\frac{x}{4}z}\Big(1-\frac{x}{4z}+\frac{x^2}{2!(4z)^2}-...\Big)\frac{dz}{z}$$
Keeping several first terms,
$$=e^\frac{x}{2}\int_{x/4}^\infty \frac{e^{-t}}{t}dt-\frac{e^\frac{x}{2}x^2}{16}\int_{x/4}^\infty\frac{e^{-t}}{t^2}dt+\frac{e^\frac{x}{2}x^2}{32}\int_1^\infty\frac{dz}{z^3}+...$$
Integrating two first terms by part,
$$I=e^\frac{x}{2}\ln t \,e^{-t}\,\bigg|_{t=x/4}^\infty+e^\frac{x}{2}\int_0^\infty\ln t\,e^{-t}dt-e^\frac{x}{2}\int_0^{x/4}\ln t\,e^{-t}dt$$
$$+\frac{e^\frac{x}{2}x^2}{16}\frac{e^{-t}}{t}\,\bigg|_{t=x/4}^\infty+\frac{e^\frac{x}{2}x^2}{16}\int_{x/4}^\infty\frac{e^{-t}}{t}dt+...$$
After straightforward rearrangement,
$$=-\ln\frac{x}{4}e^\frac{x}{4}-\gamma e^\frac{x}{2}-\frac{x}{4}\ln\frac{x}{4}e^\frac{x}{2}+\frac{x}{4}e^\frac{x}{2}-\frac{x}{4}e^\frac{x}{4}+...$$
$$\boxed{\,\,I(x)=-\ln x+2\ln2-\gamma-\frac{x}{2}\ln x+\frac{x}{2}\big(2\ln2-\gamma\big)+O\big(x^2\ln x\big)\,\,}$$
Numeric check at WolframAlpha: at $x=0.1$
$$I(x)=3.2739...$$
$$-\ln x+2\ln2-\gamma-\frac{x}{2}\ln x+\frac{x}{2}\big(2\ln2-\gamma\big)=3.2672...$$
A: Notice that $t\mapsto 1/\sqrt{t^2+t}$ behaves like $1/\sqrt{t}$ near $t=0$.
So  $1/\sqrt{t^2+t}$ is absolutely integrable on $[0,1]$, and we can use the dominated convergence theorem to compute
$$\lim_{x\rightarrow 0^+}\int_{0}^{1}e^{-xt}/\sqrt{t^2+t}dt=\int_{0}^{1}1/\sqrt{t^2+t}dt,$$
which a finite number that you can find explicitly.
Now since $t^2+t<(1+t)^2$ for $t>0$, we have
$$\int_{1}^{\infty}e^{-xt}/\sqrt{t^2+t} dt>\int_{1}^{\infty}e^{-xt}/(1+t) dt:=I.$$
Now, integrating by parts you find
$$I=\frac{1}{x}\left(\frac{e^{-x}}{2}+\int_{1}^{\infty}\frac{e^{-xt}}{(1+t)^2} dt\right)$$
But
$$\lim_{x\rightarrow 0^+}\left(\frac{e^{-x}}{2}+\int_{1}^{\infty}\frac{e^{-xt}}{(1+t)^2} dt\right)=\frac{1}{2}+\int_{1}^{\infty}\frac{1}{(1+t)^2} dt$$
by the dominated convergence theorem and the integral $\int_{1}^{\infty}\frac{1}{(1+t)^2}=\frac{1}{2}$. And of course $\lim_{x\rightarrow 0^+} \frac{1}{x}=+\infty$.
