Question

The value of $9999 \int_{0}^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^{100}}$ is _____.

Answer 1

100

Answer 2

The problem requires evaluating a definite integral involving the term $x+\sqrt{1+x^2}$. This form suggests using a hyperbolic substitution.

Let $x = \sinh t$.
Then, $dx = \cosh t \, dt$.
Also, $\sqrt{1+x^2} = \sqrt{1+\sinh^2 t} = \sqrt{\cosh^2 t} = \cosh t$ (since $x \ge 0$, $t \ge 0$, so $\cosh t > 0$).

Substitute these into the term $(x+\sqrt{1+x^2})$:
$x+\sqrt{1+x^2} = \sinh t + \cosh t = e^t$.

Now, change the limits of integration:
When $x=0$, $\sinh t = 0 \implies t=0$.
When $x \to \infty$, $\sinh t \to \infty \implies t \to \infty$.

Substitute everything into the integral $I = \int_{0}^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^{100}}$:
$I = \int_{0}^{\infty} \frac{\cosh t \, dt}{(e^t)^{100}}$
$I = \int_{0}^{\infty} \frac{\cosh t}{e^{100t}} \, dt$
$I = \int_{0}^{\infty} \cosh t \cdot e^{-100t} \, dt$

Recall the definition of $\cosh t = \frac{e^t + e^{-t}}{2}$.
$I = \int_{0}^{\infty} \frac{e^t + e^{-t}}{2} e^{-100t} \, dt$
$I = \frac{1}{2} \int_{0}^{\infty} (e^t \cdot e^{-100t} + e^{-t} \cdot e^{-100t}) \, dt$
$I = \frac{1}{2} \int_{0}^{\infty} (e^{-99t} + e^{-101t}) \, dt$

Now, integrate term by term:
$\int e^{-at} \, dt = -\frac{1}{a} e^{-at}$.
$I = \frac{1}{2} \left[ -\frac{1}{99} e^{-99t} - \frac{1}{101} e^{-101t} \right]_{0}^{\infty}$

Evaluate the definite integral using the limits:
As $t \to \infty$, $e^{-99t} \to 0$ and $e^{-101t} \to 0$.
At $t=0$, $e^{-99(0)} = 1$ and $e^{-101(0)} = 1$.

$I = \frac{1}{2} \left[ (0 - 0) - \left( -\frac{1}{99}(1) - \frac{1}{101}(1) \right) \right]$
$I = \frac{1}{2} \left[ 0 - \left( -\frac{1}{99} - \frac{1}{101} \right) \right]$
$I = \frac{1}{2} \left( \frac{1}{99} + \frac{1}{101} \right)$
$I = \frac{1}{2} \left( \frac{101 + 99}{99 \times 101} \right)$
$I = \frac{1}{2} \left( \frac{200}{9999} \right)$
$I = \frac{100}{9999}$

The question asks for the value of $9999 \int_{0}^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^{100}}$, which is $9999 I$.
$9999 I = 9999 \times \frac{100}{9999} = 100$.