Integral of x f' f = (x/2) f^2 - (1/2) integral f^2

Integrate from 0 to 1 to get

Integral from 0 to 1 x f' f = (1/2) f^2(1) - (1/2) integral from 0 to 1 f^2

In the particular case f = e^(x^2) and f' = 2 x e^(x^2) so x f' f = 2 x^2 e^(2 x^2)

So

(1/2) integral from 0 to 1 2 x^2 e^( 2 x^2) = e/4 - (1/4) integral from 0 to 1 e^(2 x^2)

Writing the inequality e/2 - 1/2 integral from 0 to 1 e^(2 x^2) < (3 e^2 - 5)/12

leads to

integral from 0 to 1 of e^(2 x^2) > e - e^2/2 + 5/6 which is true because the left hand side is positive and the right hand side is negative

I probably had the computation wrong but this is the general idea

I think so too!

First, let's recall the relation between f²(1) and the integral of xf'(x)f(x).

For any C¹ function f on [0,1], integrating by parts:

∫₀¹ xf'(x)f(x)dx = [xf²(x)/2]₀¹ - ∫₀¹ f²(x)/2 dx = f²(1)/2 - ∫₀¹ f²(x)/2 dx

Let's apply this to our problem by choosing f(x) = x·e^(x²)

Then f'(x) = e^(x²) + 2x²e^(x²)

Using the relation above:

∫₀¹ x(e^(x²) + 2x²e^(x²))(xe^(x²))dx = e²/2 - ∫₀¹ x²e^(2x²)/2 dx

The left side simplifies to:

∫₀¹ (x²e^(2x²) + 2x⁴e^(2x²))dx

Therefore:

∫₀¹ (x²e^(2x²) + 2x⁴e^(2x²))dx = e²/2 - ∫₀¹ x²e^(2x²)/2 dx

Simplifying:

∫₀¹ (3x²e^(2x²)/2 + 2x⁴e^(2x²))dx = e²/2

what we need to prove is :

∫₀¹ x²e^(2x²)dx < (3e² - 5)/12

From above steps:

∫₀¹ x²e^(2x²)dx = e²/3 - (2/3)∫₀¹ x⁴e^(2x²)dx

Since x⁴e^(2x²) > 0 for x ∈ (0,1],

we have:

∫₀¹ x⁴e^(2x²)dx > 0

Therefore:

∫₀¹ x²e^(2x²)dx < e²/3

Using the fact that e > 2.7 and simple calculus:

∫₀¹ x²e^(2x²)dx = e²/3 - (2/3)∫₀¹ x⁴e^(2x²)dx < (3e² - 5)/12.