Question

The value of the improper integral \( \int_{-\infty}^{0} \frac{e^x}{e^{2x} + 1} \, dx \) is equal to \dots\dots.

• A. \( 0 \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{4} \)
• D. \( -\infty \)

Hint:

Whenever an exponential expression like \(e^x\) appears in the numerator alongside an \(e^{2x}\) term in the denominator, substitute \(u = e^x\) to transform it into the standard arc-tangent form \(\int \frac{du}{1+u^2}\).

Answer 1

The Correct Option is C

Concept: An improper integral with an infinite limit of integration is evaluated by defining a finite boundary limit: \[ \int_{-\infty}^{0} f(x) \, dx = \lim_{a \to -\infty} \int_{a}^{0} f(x) \, dx \] To evaluate integrals featuring exponential functional forms, a substitution technique simplifies the expression into a standard derivative form.

Step 1: Apply Substitution Method
Let us define the integral: \[ I = \int_{-\infty}^{0} \frac{e^x}{(e^x)^2 + 1} \, dx \] Choose the substitution variable: \[ u = e^x \] Differentiating both sides gives: \[ du = e^x \, dx \] Now, determine the new integration limits for \( u \):
• As lower limit \( x \to -\infty \): \( u = \lim_{x \to -\infty} e^x = 0 \)
• As upper limit \( x = 0 \): \( u = e^0 = 1 \)

Step 2: Integrate with respect to the new variable \( u \)
Substitute these components back into the integral equation: \[ I = \int_{0}^{1} \frac{1}{u^2 + 1} \, du \] We recognize this as the derivative of the inverse tangent function: \[ \int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u) \] Applying the definite boundaries from 0 to 1: \[ I = \left[ \tan^{-1}(u) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) \]

Step 3: Evaluate numeric values
We know that: \[ \tan^{-1}(1) = \frac{\pi}{4} \] \[ \tan^{-1}(0) = 0 \] Thus, \[ I = \frac{\pi}{4} - 0 = \frac{\pi}{4} \] The evaluation yields \( \frac{\pi}{4} \), which matches Option (C).