We are in the right year for this! - Part 6

Calculus Level 3

$\large \int_{-\infty}^\infty e^{2016x - e^{2x}} \, dx = \, ?$

2016 None of the given choices

e^{2016}

2016 * \ln(2016)

2016!

1 solution

Arghyadeep Chatterjee
Sep 30, 2019

Let us make the substitution $e^{x} = t$ .

So the integral becomes:-

$\int_{0}^{\infty} t^{2015}e^{-t^2} dt$

Now again making a substitution of $t^2 = z$

The integral becomes:-

$\frac{1}{2}\int_{0}^{\infty}z^{1007}e^{-z} dz$

$=\frac{1}{2}\Gamma(1008)$

= $\frac{1007!}{2}$

Correct me if my answer does not match with anyone.

Arghyadeep Chatterjee - 1 year, 8 months ago