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Math  /  Calculus

Questionb) Given that ∫−∞∞e−t2dt=π\int_{-\infty}^{\infty} e^{-t^{2}} d t=\sqrt{\pi}, calculate the improper integral: ∫−∞∞xe−(x+3)2dx\int_{-\infty}^{\infty} x e^{-(x+3)^{2}} d x

Studdy Solution

STEP 1

1. The given integral ∫−∞∞e−t2dt=π\int_{-\infty}^{\infty} e^{-t^{2}} d t=\sqrt{\pi} is a known result of the Gaussian integral.
2. The integral ∫−∞∞xe−(x+3)2dx\int_{-\infty}^{\infty} x e^{-(x+3)^{2}} d x can be evaluated using symmetry and substitution techniques.

STEP 2

1. Simplify the integrand by completing the square.
2. Use substitution to transform the integral into a form involving the Gaussian integral.
3. Evaluate the transformed integral using properties of odd functions and known results.

STEP 3

First, we need to complete the square for the exponent in the integrand. The integrand is xe−(x+3)2x e^{-(x+3)^{2}}.
−(x+3)2=−(x2+6x+9) -(x+3)^2 = -\left(x^2 + 6x + 9\right)
So,
xe−(x+3)2=xe−x2−6x−9 x e^{-(x+3)^2} = x e^{-x^2 - 6x - 9}

STEP 4

Next, we substitute u=x+3u = x + 3 to simplify the integral. Let u=x+3u = x + 3, then du=dxdu = dx and x=u−3x = u - 3.
The limits of integration remain the same (−∞-\infty to ∞\infty) since they are not affected by the constant shift.
∫−∞∞xe−(x+3)2dx=∫−∞∞(u−3)e−u2du \int_{-\infty}^{\infty} x e^{-(x+3)^2} dx = \int_{-\infty}^{\infty} (u-3) e^{-u^2} du

STEP 5

Split the integral into two separate integrals:
∫−∞∞(u−3)e−u2du=∫−∞∞ue−u2du−3∫−∞∞e−u2du \int_{-\infty}^{\infty} (u-3) e^{-u^2} du = \int_{-\infty}^{\infty} u e^{-u^2} du - 3 \int_{-\infty}^{\infty} e^{-u^2} du
We know that ∫−∞∞e−u2du=π \int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi} .

STEP 6

Evaluate the first integral ∫−∞∞ue−u2du\int_{-\infty}^{\infty} u e^{-u^2} du. Note that ue−u2u e^{-u^2} is an odd function because f(−u)=−f(u)f(-u) = -f(u).
The integral of an odd function over symmetric limits around zero is zero:
∫−∞∞ue−u2du=0 \int_{-\infty}^{\infty} u e^{-u^2} du = 0

STEP 7

Evaluate the second integral:
−3∫−∞∞e−u2du=−3π -3 \int_{-\infty}^{\infty} e^{-u^2} du = -3 \sqrt{\pi}

STEP 8

Combine the results of the two integrals:
∫−∞∞(u−3)e−u2du=0−3π=−3π \int_{-\infty}^{\infty} (u-3) e^{-u^2} du = 0 - 3\sqrt{\pi} = -3\sqrt{\pi}
Solution:
∫−∞∞xe−(x+3)2dx=−3π \int_{-\infty}^{\infty} x e^{-(x+3)^{2}} dx = -3\sqrt{\pi}

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