Question

Need help , please

Answer 1

Post the question and I'll do my best :)

Answer 2

\[\int_0^{\infty}\dfrac{1}{2\sqrt{2\pi}} e^{-x/8} x^{-1/2}dx\]

Answer 3

Time to grab a pen and paper, unlike most of the questions here you can't do this one in your head lol.

Answer 4

use the gamma distribution

Answer 5

:) this is from it, I have to verify the gamma distribution by taking integral and get 1. This problem is the next step of the previous one.

Answer 6

you should verify it in general (the full gamma distribution) then plug in the numbers you get to show the integral is 1

Answer 7

as long as you know

\[\Gamma(1/2)=\sqrt{\pi|}\] the derivation is easy

Answer 8

\[\Gamma(1/2)=\sqrt{\pi}\]

some how got an extra line in there

Answer 9

You might find this interesting and fairly readable if you really want to solve the integral and not just appeal to "gamma = etc..."

http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf

The answer you're looking for comes once you get to the bottom of page 6.

Answer 10

I tried but don't get the answer :)

Answer 11

You'll have to do a substitution to get the gamma function to work out, perhaps that's where your error lies?

Answer 12

I'll show you my work, does that sound fine to you?

Answer 13

sure!!
I got Zarkon's idea. But if there is another way to learn, why not, :)

Answer 14

Well if you got it, then I mean I am just gonna do what he suggested.

Answer 15

:) To me, to this kind of problem, I will do a usual way ( substitute, by parts or else). But if there is no way to solve. I must go on his way to get the answer. However, that is kind of go around. I mean using 1 as result and prove something.

Answer 16

By definition, I have the result is 1 but the problem is asking me prove it is 1. That's the problem. :)

Answer 17

Show me your stuff, please.

Answer 18

I mean, that link should basically show you the way there for the most part. I didn't do it that way because it's too long and I've done it myself before. I just want to help out, I have a life and stuff lol.

Answer 19

lol.... It's perfectly ok, thanks for your time.

Answer 20

If you want to bypass using the Gamma function directly, you can...but you'll need to use the fact that \[\int_0^{\infty}e^{-t^2}\,dt = \dfrac{\sqrt{\pi}}{2}\] (This is known as the Gaussian integral.)

Let \(x=8t^2\implies dx=16t\,dt\). We can also change limits of integration by noting that as \(x\to 0\), \(t\to 0\) and as \(x\to\infty\), we also have \(t\to\infty\). Thus, \[\begin{aligned}\int_0^{\infty}\frac{1}{2\sqrt{2\pi}}e^{-x/8}x^{-1/2}\,dx \xrightarrow{x=8t^2}{}&\phantom{=} \int_0^{\infty}\frac{1}{2\sqrt{2\pi}}e^{-t^2}(8t^2)^{-1/2}16t\,dt\\ &= \int_0^{\infty} \frac{8(8)^{-1/2}}{\sqrt{2\pi}}e^{-t^2}t^{-1}t\,dt \\ &= \frac{\sqrt{8}}{\sqrt{2\pi}} \int_0^{\infty} e^{-t^2}\,dt \\ &= \frac{2\sqrt{2}}{\sqrt{2\pi}}\cdot\frac{\sqrt{\pi}}{2} \\ &= 1.\end{aligned}\]-----

If you still want to show more work, then you can show \(\displaystyle \int_0^{\infty}e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\) in the following way:

1. Let \(\displaystyle I=\int_0^{\infty}e^{-x^2}\,dx = \int_0^{\infty}e^{-y^2}\,dy\). Hence \(\displaystyle I^2 =\int_0^{\infty}\int_0^{\infty}e^{-(x^2+y^2)}\,dx\,dy \)

2. Convert the double integral to polar coordinates to get
\[\begin{aligned}\int_0^{\infty}\int_0^{\infty}e^{-(x^2+y^2)}\,dx\,dy &= \int_0^{\pi/2}\int_0^{\infty} re^{-r^2}\,dr\,d\theta\\ &= \left(\int_0^{\pi/2}\,d\theta\right)\left(\int_0^{\infty}re^{-r^2}\,dr\right)\end{aligned} \]

3. Evaluate the integral to get \(I^2 = \dfrac{\pi}{4}\).

4. Conclude that \(\displaystyle I=\int_0^{\infty}e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\).

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I hope this makes sense!

Answer 21

Yes, it makes perfect sense. Thank you very much.