Solve a complex improper integral involving exponential and square root functions | Step-by-Step Solution

What We're Solving

We need to evaluate the improper integral ∫[-∞ to 0] (e^(-1+4y²)/(4y)) / √(4πy) dy. This is a challenging integral that involves exponential functions, fractional powers, and an improper domain - exactly the kind that tests your understanding of advanced integration techniques!

The Approach

This integral has several red flags that suggest we need to be careful:

• 1. Complex domain: We're integrating from -∞ to 0, which includes negative values
• 2. Square root of negative numbers: √(4πy) becomes problematic when y < 0
• 3. Exponential with fractional exponent: The term e^(4y²/(4y)) = e^y needs careful handling

The key insight is that when y < 0, we're dealing with √(4πy) = √(4π|y|)·√(-1) = √(4π|y|)·i, which introduces complex numbers into our integral!

Step-by-Step Solution

Step 1: Recognize the complex nature Since we're integrating over negative y values, let's substitute y = -t where t > 0. Then:

• When y goes from -∞ to 0, t goes from +∞ to 0
• dy = -dt
• y = -t, so √(4πy) = √(4π(-t)) = √(4πt)·i

Step 2: Transform the integral ∫[-∞ to 0] (e^(-1+4y²/(4y))) / √(4πy) dy = ∫[∞ to 0] (e^(-1+4(-t)²/(4(-t)))) / (√(4πt)·i) (-dt)

Simplifying the exponent: 4(-t)²/(4(-t)) = 4t²/(-4t) = -t

Step 3: Rewrite with proper limits = ∫[0 to ∞] (e^(-1-t)) / (√(4πt)·i) dt = (1/i) ∫[0 to ∞] (e^(-1-t)) / √(4πt) dt = (-i) ∫[0 to ∞] (e^(-1-t)) / √(4πt) dt

Step 4: Factor out constants = (-i)e^(-1) ∫[0 to ∞] e^(-t) / √(4πt) dt = (-i/e) ∫[0 to ∞] e^(-t) / √(4πt) dt

Step 5: Recognize the standard integral The integral ∫[0 to ∞] e^(-t) / √(4πt) dt is a form of the Gamma function relationship: ∫[0 to ∞] e^(-t) / √(πt) dt = 1

So ∫[0 to ∞] e^(-t) / √(4πt) dt = (1/2) ∫[0 to ∞] e^(-t) / √(πt) dt = 1/2

Step 6: Complete the calculation = (-i/e) × (1/2) = -i/(2e)

The Answer

The integral evaluates to -i/(2e), confirming Wolfram Alpha's result!

Memory Tip

When you see an integral over negative values involving square roots of the variable, immediately think "complex numbers!" The appearance of i in the answer is your clue that the integral naturally lives in the complex plane. Remember: √(negative number) = i√(positive number), and this often leads to elegant results involving i in the final answer.

Great job working through this challenging problem - integrals that venture into complex analysis show you're really advancing in your mathematical journey! 🌟