Question

`\sf \left[ \: \: \int \limits_{ \sum \limits_(n = 0)^( \infty ) \big( √(n) - √(n + 1) \big)}^{ \sum \limits_(n = 0)^( \infty ) \frac{( - 1 {)}^n }{n + 1} } \left ( \lim_(t \to \infty ) \bigg(1 + \frac{1}{ {e}^(t)} \bigg )^{ {e}^(t) } \right )^{ \large(d)/(dx) \bigg( \frac{ {x}^(2) }{ {sin}^(2) x + {cos}^(2)x } \bigg)} \: dx\right]^(2) \\`​

Answer 1

The lower limit of the integral is -∞, since √n - √(n + 1) ≤ -1 for all n, and

`\displaystyle\sum_(n=0)^\infty (\sqrt n - √(n+1)) = -\sum_(n=0)^\infty\frac1\sqrt n + √(n+1)\right \ge -\sum_(n=0)^\infty \frac1{n^(1/2)}`

and the sum on the right is a divergent p-series.

The upper limit of the integral is ln(2). Recall that for |x| < 1,

`\displaystyle \sum_(n=0)^\infty x^n = \frac1{1-x}`

Integrating both sides gives

`\displaystyle \sum_(n=0)^\infty (x^(n+1))/(n+1) = -\ln(1-x) + C`

When x = 0, it follows that C = 0. As x → -1 from above, we find

`\displaystyle \sum_(n=0)^\infty ((-1)^n)/(n+1) = \ln(2)`

The limit in the integrand is e, since

`\displaystyle \lim_(t\to\infty)\left(1+\frac1{e^t}\right)^(e^t) = \lim_(n\to\infty)\left(1+\frac1n\right)^n = e`

where we replace n = eᵗ, so both n and t → ∞.

The derivative in the exponent of the integrand is

`(d)/(dx)(x^2)/(\sin^2(x)+\cos^2(x)) = (d)/(dx)x^2 = 2x`

So, the original expression simplifies significantly to

`\left(\displaystyle \int_(-\infty)^(\ln(2)) e^(2x) \, dx\right)^2`

The remaining integral is trivial:

`\displaystyle \int_(-\infty)^(\ln(2)) e^(2x) \, dx = \frac12 e^(2\ln(2)) = 2`

and so the expression has a value of 2² = 4.