Proving $ \int_1^{t}\frac{1}{x}\exp(-4(\sqrt{t+1}- \sqrt{x+1}))dx \leq \frac{c}{\sqrt{t}} $? Is there a constant $c$ such that
$$\int_1^{t}\frac{1}{x}\exp\left[-4\Big(\sqrt{t+1}- \sqrt{x+1}\Big)\right]\mathrm{d} x
\le \frac{c}{\sqrt{t}}$$
for all $t \geq 1$?
My Approach: In simulation, this holds for $c = 2$. I have tried to prove this by breaking the integral into two parts $\int_1^{\lambda t}$ and $\int_{\lambda t}^{t}$ for some $\lambda < 1$. For the first part the exponential term dominates and $1/x \leq 1$. For the second part $1/x \leq 1/\sqrt{\lambda t}$ and the integral can be calculated. Any ideas?
 A: HINT.-The answer is YES.
We have equivalently $$\frac{\sqrt t}{e^{4\sqrt{1+t}}}\int_1^t\frac{dx}{xe^{\sqrt{x+1}}}\lt c$$ and
$$\int_1^t\frac{dx}{xe^{\sqrt{x+1}}}\lt\int_1^{\infty}\frac{1}{x^2}dx=1$$
and since  $\dfrac{\sqrt t}{e^{4\sqrt{1+t}}}\lt1$ we can finish.
(one has actually $\dfrac{\sqrt t}{e^{4\sqrt{1+t}}}$ is very small and even its integral between $1$ and $\infty$ is approximately equal to $0.003968$).
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I have made a slip, noticed by River Li, and, by regulation, I have not been able to delete my answer as I wanted because it has already been accepted.
I have no willingness now to work with this problem in detail but I can give the following HINT: we can reduce the problem to a single function $F(x)=f(x)g(x)$ where
$$f(x)=\frac{\sqrt x}{e^\sqrt {x+1}}\\g(x)=\int_1^x\frac{e^\sqrt{t+1}}{t}dt$$
Now every numerical value of $F(a)$ for $a\ge 1$ is very small (despite the divergence of $g(x)$ and due to the fast convergence to $0$ of $f(x)$ and it can be ensured that
$F (x)\le 0.005$ (but this value of the constant $ c $ is not the best)
A: Too long for comments.
This is a very interesting problem from a numerical point of view. Sooner or later, you will learn about the exponential integral function.
Without entering in the details (available if you wish), let $f(t)$ to be the result of the integral. Now, consider the objective function
$$\Phi(t,c)=\left(f(t)-\frac{c}{\sqrt{t}}\right)^2$$ to be minimized with respect to $t$ and $c$; this means that we look for the points where the two curves are tangent to eachother.
Numerically, this happens at $t=4.1636$ and $c=0.654764$. So, the inequality holds as soon as $c \geq 0.655$.
A: $$I=e^{-4\sqrt{t+1}}\int_1^t\frac{e^{4\sqrt{x+1}}}{x}dx$$
the integrand is:
$$f(x)=\frac{e^{4\sqrt{x+1}}}{x}$$
and we know that:
$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$
so for a positive $x$ (which we have here) we can say that:
$$e^x<\sum_{n=0}^N\frac{x^n}{n!},N<\infty$$
so for our:
$$e^{4\sqrt{x+1}}<1+4\sqrt{x+1}$$
so we can say that within our domain:
$$f(x)<\frac{1+4\sqrt{x+1}}{x}$$
since this function is decreasing for an increasing $x$ i.e. $x\in[1,t],f'(x)<0$
we know that the minimum is at $f(t)$ and the maxima is at $f(1)$, using these to approximate the integral.

Alternatively, looking at the function:
$$x\in[1,t],g(x)=\sqrt{t+1}-\sqrt{x+1}$$
we notice that $g(x)\ge0\therefore \exp(-4g(x))\le1$
