- #1

- 613

- 1

## Main Question or Discussion Point

Hey guys

I've been reading through a few books and I can't seem to work this out;

[tex]\int _0^{\infty }e^{-a x^2}xdx = \frac{1}{2a}[/tex]

I keep getting

[tex]\int _0^{\infty }e^{-a x^2}xdx = -\frac{1}{2a}[/tex]

I do the old variable switch

[tex]u = x^2[/tex]

[tex]du=2 x dx[/tex]

Which leaves me with

[tex]\int_0^{\infty } \frac{e^{-a u}}{2} \, du[/tex]

Which them leads me to

[tex]\text{Lim} u\to 0\frac{e^{- a u}}{-2 a}=-\frac{1}{2a}[/tex]

Where am I picking up this unwanted - from?

I can't see where I've gone wrong

I've been reading through a few books and I can't seem to work this out;

[tex]\int _0^{\infty }e^{-a x^2}xdx = \frac{1}{2a}[/tex]

I keep getting

[tex]\int _0^{\infty }e^{-a x^2}xdx = -\frac{1}{2a}[/tex]

I do the old variable switch

[tex]u = x^2[/tex]

[tex]du=2 x dx[/tex]

Which leaves me with

[tex]\int_0^{\infty } \frac{e^{-a u}}{2} \, du[/tex]

Which them leads me to

[tex]\text{Lim} u\to 0\frac{e^{- a u}}{-2 a}=-\frac{1}{2a}[/tex]

Where am I picking up this unwanted - from?

I can't see where I've gone wrong