1

Could someone explain why the following change of variable is valid?:

$$2 \int^{\sqrt t}_0 \frac 1 {\sqrt {2\pi}} e^{-u^2/2} du = \frac 1 {\sqrt {2 \pi}} \int^t_0 e^{-s/2}s^{-1/2}ds$$

Using the substitution method I've been used to, I can write $s = u^2$ which implies $\frac {ds} {du} = 2u$. So $ds = 2u (du)$. But $2u (du)$ is not to be found in the integral, so obviously $ds/u$ is substituted for $2(du)$. However why is this permissible ?

Shuzheng
  • 5,533

1 Answers1

2

You need to change $du$ to $ds$...

You got $$\frac{ds}{du}=2u.$$

Then, what we want is to get the following form : $$du=F(s)ds$$ for a function $F$.

Since $$s=u^2,s\ge0, u\ge0,$$ we have $$u=\sqrt s=s^{1/2}.$$ So, we have $$\frac{ds}{du}=2u=2s^{1/2}.$$

Hence, the first integral will be $$2\int_{0}^{t}\frac{1}{\sqrt{2\pi}}e^{-s/2}\cdot \frac{1}{2s^{1/2}}ds.$$

EDIT : $$\int_{g(0)}^{g(t)}f(u)du=\int_{0}^{t}f(g(s))g^\prime(s)ds$$ where $$f(u)=\frac{2}{\sqrt{2\pi}}e^{-u^2/2},g(s)=\sqrt s.$$

mathlove
  • 139,939
  • You are welcome! my pleasure. – mathlove Jan 03 '14 at 11:53
  • Sorry to bother you once more, but why dont we write $du = 2s^{1/2}$ then we get $\int f(s) \cdot ds/du = F(s)$ (right)? - and then evaluate for new limits. We are we multiplying by $(ds/du)^{-1}$ ? – Shuzheng Jan 03 '14 at 12:12
  • No, it is not correct. You want to change a variable $u$ to $s$, right? we have $ds/du=2u.$ so, how can we get $du=2s^{1/2}$? Remember that $ds$ and $du$ have to be in the same equation. – mathlove Jan 03 '14 at 12:19
  • We get $ds/du = 2s^{1/2}$. So now our integral is in the form $\int f(s) \cdot \frac 1 {2s^{1/2}} ds$. To to use our rule $\int f(g(x)) \cdot g'(x) = F(g(x))$ we must have $\frac 1 {2s^{1/2}} = ds/du$, which it is not ? – Shuzheng Jan 03 '14 at 12:44
  • We have $ds/du=2s^{1/2}\iff du=ds/2s^{1/2}.$ so, you can set this into the first integral. This is all we do. – mathlove Jan 03 '14 at 14:40
  • But we cannot calculate with $ds/du$ like it was a fraction. These are only symbols. We must bring the integral into the form shown above ? If not, please tell how you justify your manipulation. – Shuzheng Jan 03 '14 at 14:43
  • A good point. see here. You'll learn a lot. http://math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio – mathlove Jan 03 '14 at 14:51
  • So we are not doing regular substitution rule like the one I'm mentioning above ? And does your method has a name ? – Shuzheng Jan 03 '14 at 14:54
  • Technically, we can 'set $du=ds/2s^{1/2}$ in the first integral'. This is all we do. This is not substitution. We just change a variable from $u$ to $s$. In the process to change it, we can treat $du=ds/2s^{1/2}$ 'as a fraction'. This is how we solve integrals. So, I think first you need to know how to solve it, and know why we can do like this. In short, you cannot find the value of integrals in your way. Just learn and try it. – mathlove Jan 03 '14 at 15:02
  • It makes sense, although I've never solved integrals like this. In Calculus courses, I've only worked with well-known rules like integration by substitution and parts. However where can I read about your method formally ? Does this method has a name ? Where is it proved ? – Shuzheng Jan 03 '14 at 15:05
  • Well, many many textbooks... You can go to library, or book shop or internet... Just find information. There are tons of sites. so, google "substitution method" (I said this is not substitution, which means your way is not substitution, sorry it may confuse you) I think you don't really understand how to use substitution method. Just read a book or a site carefully and try to solve an easy quetion...Good luck! – mathlove Jan 03 '14 at 15:11
  • Well, it requires a proof that you can substitute $du = ds/F(s)$ for some function $F$, change the limit of the integral and then everything works. The substitution method is proved rigorous here: http://en.wikipedia.org/wiki/Integration_by_substitution Just because one solves some questions using your strategy works doesn't mean it is true. – Shuzheng Jan 03 '14 at 15:14
  • Sorry, I don't get what you really want to say. – mathlove Jan 03 '14 at 15:16
  • Is this the version of integration by substitution you are using ?: http://en.wikipedia.org/wiki/Integration_by_substitution. If yes, can you write our integral as $\int^a_b f(g(x))g'(x) dx$ and apply this integration rule to solve the problem ? If this is not the integration rule you are using, please provide a link to it - just a single webpage where it is stated formally. – Shuzheng Jan 03 '14 at 15:29
  • @user111854 : see an edit in my answer. – mathlove Jan 03 '14 at 15:42
  • Yes, this is totally right. We are using the substitution rule in reverse to get the answer. How come the other way of doing it provide the same answer ? – Shuzheng Jan 03 '14 at 15:50
  • What is 'the other way'? – mathlove Jan 03 '14 at 15:52