# Improper Integrals

• Jan 24th 2007, 04:18 PM
turtle
Improper Integrals
Does anyone know how to do these problems. I'm stuck on them, and I don't know where to start. Thanks for any help.

1. the integral from 0 to 1 of (theta + 1)/ the square root of ( theta^2 + 2 theta)

2. the integral from 0 to infinty of (16tangent inverse of x)/(1 + x^2) dx

3. the integral from negative infinty to infinity of 2xe^(-xsquared) dx
• Jan 24th 2007, 04:48 PM
ThePerfectHacker
Quote:

Originally Posted by turtle
1. the integral from 0 to 1 of (theta + 1)/ the square root of ( theta^2 + 2 theta)

$\displaystyle \int_0^1 \frac{x+1}{\sqrt{x^2+2x}} dx$
This is a type two improper integral for it is not defined at a point in the interval.

We will find,
$\displaystyle \lim_{t\to 0^+} \int_t^1 \frac{x+1}{\sqrt{x^2+2x}} dx$
You can find,
$\displaystyle \int \frac{x+1}{\sqrt{x^2+2x}} dx$
By using the inner function,
$\displaystyle u=x^2+2x$
Thus,
$\displaystyle \frac{1}{2} \int \frac{1}{\sqrt{u}} du=\frac{1}{2} \cdot \frac{2}{1} u^{1/2} +C=\sqrt{x^2+2x}+C$
Thus,
$\displaystyle \lim_{t\to 0^+}\sqrt{1^2+2(1)}-\sqrt{t^2+2t}=\sqrt{3}$
• Jan 24th 2007, 04:52 PM
ThePerfectHacker
Quote:

Originally Posted by turtle

2. the integral from 0 to infinty of (16tangent inverse of x)/(1 + x^2) dx

$\displaystyle \int_0^{\infty} \frac{16 \tan^{-1} x}{1+x^2} dx$
To find,
$\displaystyle \int \frac{16 \tan^{-1} x}{1+x^2} dx$
Use the inner function $\displaystyle u=\tan^{-1} x$.
To get,
$\displaystyle \int 16 u du=8u^2+C=8(\tan^{-1} x)^2 +C$
Thus, the integral is,
$\displaystyle \lim_{t\to \infty}8 (\tan^{-1} t)^2 - 8(\tan^{-1} 0)^2=8\cdot \frac{\pi^2}{4}=2\pi^2$
• Jan 24th 2007, 04:56 PM
ThePerfectHacker
Quote:

Originally Posted by turtle
3. the integral from negative infinty to infinity of 2xe^(-xsquared) dx

Subdivide,
$\displaystyle \int_{-\infty}^0 2xe^{-x^2} dx+\int_0^{\infty} 2xe^{-x^2}dx$
The integral,
$\displaystyle \int 2xe^{-x^2} dx=e^{-x^2}+C$
Thus,
$\displaystyle \lim_{t\to -\infty}e^{-0^2}-e^{-t^2}+\lim_{t\to \infty} e^{-t^2}-e^{-0^2} =0$
Of you can use the fact that,
$\displaystyle y=2xe^{-x^2}$ is odd.
• Jan 24th 2007, 05:05 PM
Soroban
Hello, turtle!

These require only simple substitutions . . .

Quote:

$\displaystyle 1)\;\;\int^1_0\frac{\theta + 1}{\sqrt{\theta^2 + 2\theta}}\,d\theta$

We have: .$\displaystyle \int(\theta^2 + 2\theta)^{-\frac{1}{2}}(\theta + 1)\,d\theta$

Let: $\displaystyle u = \theta^2 + 2\theta\quad\Rightarrow\quad du = (2\theta + 2)d\theta\quad\Rightarrow\quad d\theta = \frac{du}{2(\theta + 1)}$

Substitute: .$\displaystyle \int u^{-\frac{1}{2}}(\theta + 1)\frac{du}{2(\theta + 1)} \;=\;\frac{1}{2}\int u^{-\frac{1}{2}}\,du \;=\;u^{\frac{1}{2}} \;=\;\sqrt{u}$

Back-substitute: .$\displaystyle \sqrt{\theta^2 + 2\theta}\,\bigg|^1_0\;=\;\sqrt{1 + 2} - \sqrt{0 + 0} \:=\:\sqrt{3}$

Quote:

$\displaystyle 2)\;\;\int_0^{\infty} \frac{16\tan^{-1}x}{1 + x^2}\,dx$

We have: .$\displaystyle 16\int^{\infty}_0\tan^{-1}x\cdot\frac{dx}{1 + x^2}$

Let $\displaystyle u = \tan^{-1}x\quad\Rightarrow\quad du = \frac{dx}{1 + x^2}$

Substitute: .$\displaystyle 16\int^{\infty}_0u\:du \;=\;8u^2$

Back-substitute: .$\displaystyle 8\left[\tan^{-1}x\right]^2\,\bigg|^{\infty}_0$

Evaluate: .$\displaystyle 8\bigg[\left(\tan^{-1}(\infty)\right)^2 - \left(\tan^{-1}(0)\right)^2\bigg]\;=\;8\bigg[\left(\frac{\pi}{2}\right)^2 - 0^2\bigg]\;=\;2\pi^2$

Sheesh! . . . Too slow again . . .