E and pi?

Calculus Level 4

$\large \int _{ 0 }^{ \infty }{ \dfrac { x-1 }{ { e }^{ x^{ 2 }-1 } } \, dx } =\dfrac { 1 }{ A } (Be-Ce\sqrt { D\pi } )$

If the equation above holds true for positive integers $A,B,C$ and $D$ , find $A+B+C+D$ .

The answer is 5.

1 solution

Aditya Narayan Sharma
May 10, 2016

Rewrite the integral as : $\large \displaystyle S=e\int_{0}^{\infty}e^{-x^2}xdx-e\int_{0}^{\infty}e^{-x^2}dx$

$\large \displaystyle S=\frac{e}{2} \int_{0}^{\infty} e^{-x^2}d(x^2)-e\frac{\sqrt{\pi}}{2}$

$\large \displaystyle S=\frac{e}{2}-\frac{e\sqrt{\pi}}{2}$ , So the answer is : $\boxed{2+1+1+1=5}$

It is just one of the possible answers. A = 2 * B, 2 * C * sqrt(D) = A = 2 * B, lets D = [n ^ 2], where n is any positive integer, then 2 * C * n = A = 2 * B Then A + B + C + D -> 2 * C * n + C * n + C + n ^ 2 = C * (3 * n + 1) + n ^ 2. There is no way to determine dependency between C and k. So, the answer would be [m * (3 * n + 1) + n ^ 2], where n and m any positive integers. You answer (5) is for n = 1 and m = 1, however, 9 is also correct answer (m = 2, n = 1, and A = 4, B = 2, C = 2, D = 1), and 11 (m = 1, n = 2, and A = 4, B = 2, C = 1, D = 4), and 18 ( n = 2, m = 2, A = 8, B = 4, C = 2, D = 4), and so on.

Volodymyr Prokurashko - 5 years ago

Generally , we minimize the values . As you say we can never have the values as they can be changed into alternate forms but here we are to deal with the minimized values

Aditya Narayan Sharma - 5 years ago

E and pi?

The answer is 5.

1 solution

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