What is the Integral of e^(-bx^2) Using Substitution?

[orange]

Hello!

I've got a problem I've been working on for hours.

I get a clue;

If the integral (from zero to infinity) of e^(-x^2) is sqrt(pi)/2, what is
the integral (from zero to infinity) of e^(-bx^2)?

I've tried substitution, but I kind of got it wrong. If x = y/sqrt(b), I get the same integral as in the clue. But then I'm stuck with a 1/sqrt(b) which I can't get rid of. Anyone up for the challenge? Thanks..

 

[StatusX]

What do you mean get rid of? sqrt(b) is a constant, and it will appear in the final answer.

 

[orange]

What do you mean get rid of? sqrt(b) is a constant, and it will appear in the final answer.

But is sqrt(b) a constant when it is the same thing as y/x? Can I move it outside the integral?

 

[orange]

I'll take the whole problem:

The integral (from zero to +infinity) of c*x^2*e^(-bx^2) dx = 1

I get the clue: integral (from zero to +infinity) of e^(-x^2) = sqrt(pi) / 2

What is c?

 

[George Jones]

You know

\frac{\sqrt{\pi}}{2} = \int_{0}^{\infty} e^{-x^{2}} dx.

Use the substitution u = \sqrt{b} x to calculate

I \left( b \right) = \int_{0}^{\infty} e^{-bx^{2}} dx

for any b. Then differentiate with respect to b both sides of

I \left( b \right) = \int_{0}^{\infty} e^{-bx^{2}} dx

to find the integral that you want.

Regards,
George

 

[orange]

You know

\frac{\sqrt{\pi}}{2} = \int_{0}^{\infty} e^{-x^{2}} dx.

Use the substitution u = \sqrt{b} x to calculate

I \left( b \right) = \int_{0}^{\infty} e^{-bx^{2}} dx

for any b. Then differentiate with respect to b both sides of

I \left( b \right) = \int_{0}^{\infty} e^{-bx^{2}} dx

to find the integral that you want.

Regards,
George

Thanks a lot George and StatusX. Appreciate you taking your time. :!)