Question

Evaluate, ∫-[infinity][infinity] d x e^-|x+1| x sin (x)

Answer 1

The integral
`\(\int_(-\infty)^(\infty) dx \, e^(-|x+1|) x \sin(x)\)` evaluates to 0.

Step-by-step explanation:

To evaluate the given integral, we break it into two parts based on the absolute value term. The integral can be expressed as the sum of two integrals:
`\(\int_(-\infty)^(-1) e^(x+1) x \sin(x) \, dx + \int_(-1)^(\infty) e^(-x-1) x \sin(x) \, dx\).` However, the integrands in both parts are odd functions over symmetric intervals, resulting in their values canceling each other out when integrated over the entire range. Therefore, the total integral is equal to zero.

Mathematically, this cancellation occurs due to the symmetry properties of the sine function and the product of an odd function
`(\(x\))`and an even function
`(\(e^(-|x+1|)\)).` When integrated over intervals with the appropriate symmetry, the contributions from each side cancel out, leading to a net result of zero.

In summary, the integral
`\(\int_(-\infty)^(\infty) dx \, e^(-|x+1|) x \sin(x)\)`evaluates to zero. This result arises from the cancellation of contributions due to the symmetry properties of the integrands, demonstrating the importance of recognizing the characteristics of functions when evaluating integrals over infinite intervals.