Improper Integral Help

**masteroc** · November 6th 2011, 08:26 PM

Just needed a bit of confirmation that I was doing this problem correctly. I started off using a table definition that you can see at the bottom of that picture. The part that got cut off just shows: u/a^2sqrt(u^2+a^2). After that I just put in 0 and infinity, and came out with the constants * 1/r^2. Does that seem like an appropriate answer?

Thanks!

Link: imgur: the simple image sharer

**Prove It** · November 6th 2011, 08:41 PM

Originally Posted by masteroc

Just needed a bit of confirmation that I was doing this problem correctly. I started off using a table definition that you can see at the bottom of that picture. The part that got cut off just shows: u/a^2sqrt(u^2+a^2). After that I just put in 0 and infinity, and came out with the constants * 1/r^2. Does that seem like an appropriate answer?

Thanks!

Link: imgur: the simple image sharer

First you need to find the indefinite integral, then see if it converges.

Let $\displaystyle x = r\tan{\theta} \implies dx = r\sec^2{\theta}\,d\theta$ and the integral becomes

$\displaystyle \begin{align*}\int{\frac{dx}{(r^2 + x^2)^{\frac{3}{2}}}} &= \int{\frac{r\sec^2{\theta}\,d\theta}{[r^2 + (r\tan{\theta})^2]^{\frac{3}{2}}}} \\ &= \int{\frac{r\sec^2{\theta}\,d\theta}{[r^2 + r^2\tan^2{\theta}]^{\frac{3}{2}}}} \\ &= \frac{r\sec^2{\theta}\,d\theta}{[r^2(1 + \tan^2{\theta})]^{\frac{3}{2}}} \\ &= \frac{r\sec^2{\theta}\,d\theta}{(r^2\sec^2{\theta} )^{\frac{3}{2}}} \\ &= \frac{r\sec^2{\theta}\,d\theta}{r^3\sec^3{\theta}} \\ &= \int{\frac{d\theta}{r^2\sec{\theta}}} \\ &= \frac{1}{r^2}\int{\cos{\theta}\,d\theta} \\ &= \frac{\sin{\theta}}{r^2} + C \\ &= \frac{\tan{\theta}}{r^2\sqrt{1 + \tan^2{\theta}}} + C \\ &= \frac{\frac{x}{r}}{r^2\sqrt{1 + \frac{x^2}{r^2}}} + C \\ &= \frac{x}{r^3\sqrt{\frac{r^2 + x^2}{r^2}}} + C \\ &= \frac{x}{r^2\sqrt{r^2 + x^2}} + C\end{align*}$

So now to take the definite integral between $\displaystyle 0$ and $\displaystyle \infty$ , we need to evaluate

$\displaystyle \lim_{\epsilon \to \infty}\left(\frac{\epsilon}{r^2\sqrt{r^2 + \epsilon^2}}\right) - \frac{0}{r^2\sqrt{r^2 + 0^2}}$

**masteroc** · November 6th 2011, 08:51 PM

Thanks for the reassurance, I got that answer using the tables definition and when I went to evaluate, I got the right part of that (the non-limit part) equaling zero. However, the limit part i'm confused with. The top part is infinity and so is the bottom it would seem. I used LHopitals Rule, but it seemed to keep on going forever because the bottom always equaled infinity which would make this problem 0-0=0....and that does not seem correct given that this is a physics formula and usually would have an answer greater than 0.

Thanks again for the quick response!

**Prove It** · November 6th 2011, 09:19 PM

Originally Posted by masteroc

Thanks for the reassurance, I got that answer using the tables definition and when I went to evaluate, I got the right part of that (the non-limit part) equaling zero. However, the limit part i'm confused with. The top part is infinity and so is the bottom it would seem. I used LHopitals Rule, but it seemed to keep on going forever because the bottom always equaled infinity which would make this problem 0-0=0....and that does not seem correct given that this is a physics formula and usually would have an answer greater than 0.

Thanks again for the quick response!

L'Hospital's Rule isn't needed...

$\displaystyle \begin{align*} \frac{\epsilon}{r^2\sqrt{r^2 + \epsilon^2}} &= \frac{\epsilon \cdot \frac{1}{\epsilon}}{r^2\sqrt{r^2 + \epsilon ^2}\cdot \frac{1}{\epsilon}} \\ &= \frac{1}{\frac{r^2\sqrt{r^2 +\epsilon ^2}}{\sqrt{\epsilon^2}}} \\ &= \frac{1}{r^2\sqrt{\frac{r^2 + \epsilon ^2}{\epsilon ^2}}} \\ &= \frac{1}{r^2\sqrt{\frac{r^2}{\epsilon^2} + 1}} \end{align*}$

Now make $\displaystyle \epsilon \to \infty$ ...

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