## swissgirl 2 years ago Calculus: Find K

1. swissgirl

|dw:1353910548451:dw|

2. swissgirl

You need to use integration by parts

3. hartnn

put u= -y^2/2 then.

4. ivanmlerner

Is it:$\int\limits\limits_{-\infty}^{\infty}ke^{-\frac{y^2}{2}}dy=1$?

5. swissgirl

Yes

6. hartnn

u= -y^2/2 find dy =... ?

7. swissgirl

I used u=y^2

8. swissgirl

but maybe its easier with u=-y^2/2

9. swissgirl

Its not smart to have that negative there

10. swissgirl

Do you agree?

11. swissgirl

@hartnn ?

12. hartnn

oh, sorry.... yeah, its easier with e^u constants can be rearranged easily...

13. ivanmlerner

I got the wrong answer when evaluating the integral, and I won't do it again now. But this function is a gaussian, and it is expected that the constant k in this case has the value 1/sqrt(2pi)=sqrt(2pi)/2pi. That comes from the statistic theory.

14. swissgirl

ohhhhh

15. swissgirl

16. hartnn

did u try that substitution and then integration by parts ?

17. swissgirl

I was getting weird answers. I have not really touched calculus for over a yr so i cld be messing up

18. hartnn

$$\huge\int \frac{e^{-u}u^{-1/2}}{\sqrt2}du$$ when 2u=y^2 yes, its messy...

19. swissgirl

How would you continue from here?

20. swissgirl

Can you show me?

21. swissgirl

ok dont show me

22. hartnn

u need to know gamma function to solve that ... i am so sorry, my internet are giving me connection problems....

23. swissgirl

hmmm I know the gamma distribution function

24. swissgirl

ohhhh I see it now lol

25. hartnn

$$\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} dt$$

26. swissgirl

Yesssssss

27. hartnn

now , you'll get it easily.

28. hartnn

29. swissgirl

Thanks let me try to work it out then. Thanks :)))))

30. swissgirl

what?

31. hartnn

$$\Gamma{1/2} = \sqrt{\pi}$$

32. hartnn

you'll need it.

33. swissgirl

ohhh coool never knew that

34. swissgirl

I feel like i am missing the basics in math :(

35. hartnn