I have to find a equivalent when $x$ comes to $\infty$ for all $a$ (fixed) in $\mathbb{R}_+^*$ of this integral :
$$ \int_0^a \frac{e^{-xt}}{\sqrt{a-t}}\mathrm{d}t $$
My work :
For $x \in \mathbb{R}_+^*$ fixed, $$f:t\mapsto \frac{e^{-xt}}{\sqrt{a-t}}$$ is a continuous function on $(0,a[$ and $f(t)\underset{t \to a}{\sim} =\frac{e^{-xa}}{\sqrt{a-t}} $ and by Riemann and theorem about equivalence for positive functions then $f$ is integrable so we have the existence of $$I(x)=\int_0^a \frac{e^{-xt}}{\sqrt{a-t}}\mathrm{d}t$$
$$\therefore$$
Now I try this change of variable (sorry for my bad english) :
$u=\sqrt{a-t}$, so $t=a-u^2$ and $\mathrm{d}t=-2u\mathrm{d}u$ I find $$I(x)=2 e^{-ax}\cdot \int_0^{\sqrt{a}}e^{xu^2}\mathrm{d}u$$
Well I don't if I am on a good way but I have no idea yet to continue, can you help me please?
Thank you in advance
That's all I find it :D
So in the last integral I do $v=u\sqrt{x}$ then I find $$I(x)=\frac{2e^{-ax}}{\sqrt{x}}\int_0^{\sqrt{ax}} e^{v^2}\mathrm{d}v$$
Then $h : v \mapsto \frac{e^{v^2}}{2v}$ is $C^1$ on $]0,\infty[$ and $$h'(v)=e^{v^2}-h(v)$$
Then we find that $$\int_1^x e^{v^2}\mathrm{d}v \underset{x \to +\infty}{\sim} \int_1^x h'(v)\mathrm{d}v\underset{x \to +\infty}{\sim}\frac{e^{x^2}}{2x}$$
And so $$I(x)=\frac{2e^{-ax}}{\sqrt{x}}\cdot\frac{e^{ax}}{2\sqrt{ax}}=\frac{1}{x\sqrt{a}}$$
Quod Erat Demonstrandum !!!