## Substitution in integral

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 cant get rid of. Anyone up for the challenge? Thanks..

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

 Quote by StatusX 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?

## Substitution in integral

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?

 Mentor 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

 Quote by 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
Thanks alot George and StatusX. Appreciate you taking your time.