Question

Evaluate \[\large \int \limits_0^{\infty}\frac{e^{-x}}{\sqrt{x}} \, dx = \sqrt{\pi} \]
@Marki

Answer 1

i found the answer online but i don't get one step, so I won't bring it up :P

Answer 2

ok so far i got 2 methods :)

Answer 3

\[\int\limits_{ }^{ } \frac{e^{-x}}{\sqrt{x}}dx\]\[2\sqrt{x}e^{-x}+2\int\limits_{ }^{ }\sqrt{x}e^{-x}~dx\](differentiated e^-x, integ. by parts)
\[2\sqrt{x}e^{-x}+2\int\limits_{ }^{ }\sqrt{x}e^{-x}~dx\]u sub,

solving the second integral.

u=sqrt{x}
-x=-u^2
du= 1/(2sqrt{x}) dx >> 2sqrt{x}du=dx >> 2u du=dx

2 * integ. 2u^2 e^(-2u) du

integration by parts twice to reduce the u^2.

Answer 4

Laplace :3

Answer 5

my method, if I haven't made a mistake, is not hard...

Answer 6

Oh nice i have searched a bit in MSE but couldnt find this :O
please provide the link @TuringTest :)

Answer 7

Interesting xD
f(t) = 1/sqrt(t) and we want L(1) @hartnn ?

@((*-*)) im still going thru your method

Answer 8

take your time

Answer 9

yes ganesh, thats the easiest way :P

Answer 10

http://math.stackexchange.com/questions/215352/why-is-gamma-left-frac12-right-sqrt-pi
(I am looking at the one using spherical coordinates)
I don't get why the bounds change on the u-sub, and why the u-sub is correct
shouldn't it lead to 4x in the exponent?

Answer 11

I think, it is necessary integrate by parts, first, as below:
\[e ^{-x}*2\sqrt{x}|_{0}^{+\infty}+2\int\limits_{0}^{+\infty}e ^{-x}\sqrt{x}dx=2\Gamma(3/2)=2*\frac{ \sqrt{\pi} }{ 2 }=\sqrt{\pi}\]

Answer 12

lets take indefinite integral as refrence
\(\large \int \limits_{-\infty}^{\infty} {e^{-x^2}} \, dx
\\=\large 2\int \limits_0^{\infty} {e^{-x^2}} \, dx \)

x= sqrt t
so we would have
\(\large 2\int \limits_0^{\infty} {e^{-x^2}} \, dx =\large 2\int \limits_0^{\infty} \frac{1}{2} {e^{-t}} t^{-\frac{1}{2} } \, dx = \Gamma(1/2) \)

now lets proof
\(\large \int \limits_{-\infty}^{\infty} {e^{-x^2}} \, dx \)

there is several way chose from here to make me explain :P

http://en.wikipedia.org/wiki/Gaussian_integral

Answer 13

u know that definite integral from -infinity to infinity give -sqrt pi
now only chose any proof in wiki link so i would explain ,till then i'll post next method

Answer 14

sorry indefinite*

Answer 15

that looks very neat xD
I can work \(\int_0^{\infty} e^{-x^2}dx\) using polar coordinates..

lm going through that mse link provided by TT
surprisingly nobody used the obvious laplace transform there.. is it easy to find the laplace transform of 1/sqrt(t) ?

Answer 16

@hartnn mentioned that laplace was the easy way

Answer 17

@Michele_Laino how did you get \(\Gamma (3/2) = \sqrt{\pi}/2\)
only if we can get some integer after doing by parts n times xD

Answer 18

hmm i never used or liked laplace :P
my second method is using gama function properties
im just writing my notes now , will provide u with it quick

Answer 19

I was like, hey guys, how about an elementary calculus?

Answer 20

@ganeshie8 what I don't understand is why it changes to\(\int_{-\infty}^\infty e^{-x^2}dx\) and *not* \(\int_{0}^\infty e^{-x^2}dx\)
could somebody explain that to me? I am dumb

Answer 21

since for a property of the Gamma function, we can write:
\[\Gamma (z+1)=z \Gamma (z)\]
so:
\[\Gamma(3/2)=\Gamma (1+1/2)=\frac{ 1 }{ 2 }\Gamma \left( \frac{ 1 }{ 2 } \right)\]
@ganeshie8

Answer 22

well know that
\(\Gamma (p)=p\Gamma (p-1)\)
so
3/2=1+1/2
\(\Gamma (3/2)=3/2\Gamma (1/2)\)
hmm how did u got sqrt pi /2 ?

Answer 23

oh wait my bad :P
its p-1 gamma (p-1)

Answer 24

then its true :O

Answer 25

my questions got ignored :P
fine, I'll figure it out myself

Answer 26

hey no im still thinking.. it has to do with e^(-x^2) being an converging even function i think

Answer 27

hm i didnt think of that

Answer 28

\[2\int\limits_0^{\infty} e^{-x^2} dx =\int\limits_{-\infty }^{\infty} e^{-x^2} dx \]

Answer 29

specifically\[\int_0^\infty e^{-x}d(2\sqrt x)=\int_{-\infty}^\infty e^{-u^2}du\]

Answer 30

ok first of all where did the 2 go?

Answer 31

sorry @Marki
I already deleted my post

Answer 32

well ,this integral is the base of all integrals in complex :P
i wuld say its like number 1 in multiplying operation and @Michele_Laino i note that thank u :)
, how ever @TuringTest