Numerical integration of 1/((1+x)√x) from 0 to infinity

[Dominguez Scaramanga]

Hello there, I've not been here in a while, but I'm stuck doing this integration and wondered if some of you kind people would help

$$\int_0^\infty \frac{1} {(1+x)\sqrt{x}} dx$$

(appologies for the lack of spacing in there...)

anyways, I know that when x tends to infinity, the integral can be approximated to,


$$\int_0^\infty \frac{1} {(x)\sqrt{x}} dx$$

but I can't seem to find this identity in any of my tables anywhere...

The reason I need it is because I'm in the processes of writing some c code to analytically calculate this with a specified degree of acuracy, (am going to use the trapezium method of integration I think) so it would be nice to know if the answers I get out of it are any good or not!

thanks for you time

 

[DeadWolfe]

use rules of exponents.

1/(x(sqrtx)) = x^(-3/2)

 

[Dominguez Scaramanga]

awesome, thanks DeadWolfe :)

 

[dextercioby]

That integral (the first) can be very simply computed analytically to yield the result $\pi$...Heck,u can even define $\pi$ by it

$$\pi=:\int_{0}^{+\infty} \frac{dx}{(1+x)\sqrt{x}}$$

HINT:Make the obvious substitution
$$\sqrt{x}=t$$

Daniel.

 

[Dominguez Scaramanga]

wow, even more helpful, thanks a lot

also, would I be correct in saying that the limit of the integrand as x-->infinity is pi, in that case?
or have I got completely mudled up?

 

[dextercioby]

Define

$$P(x)=:\int_{0}^{x} \frac{dt}{(1+t)\sqrt{t}}$$

Show that

$$P(x)=2\arctan x$$

Then it's easy to say

$$\lim_{x\rightarrow +\infty} P(x)=\pi$$

Not the integrand!The integrands's (inferior) limit to $0$ is $+\infty$,while its limit to $+\infty$ is $0$ ()

Daniel.

 

[Dominguez Scaramanga]

thanks very much for your help dextercioby, it'll be most useful!

now all I've got to do is figure out how to do this with C

also, where you have,

$$P(x)=:\int_{0}^{x} \frac{dt}{(1+t)\sqrt{t}}$$

that'll yield the same result as for

$$F(x)=:\int_{0}^{+\infty} \frac{dx}{(1+x)\sqrt{x}}$$

with the substitution

$$\sqrt{x}=t$$

right?

 

[dextercioby]

Nope.The second (the one with F(x)) is a number,while the first is a function...So in the second case,the notation is incorrect...

Daniel.

 

[Dominguez Scaramanga]

ah I see, ok, so the first one, with x = +infinity, will equal pi, the same as the bottom one would (had I wrote it correctly).

 

[dextercioby]

Yes.U got it

Daniel.

 

[Dominguez Scaramanga]

finally thanks very much for your help