- #1

- 2

- 0

I was trying to change it into a product of an error function and a gamma function, but I needed an extra dx. Any other ideas?

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- Thread starter aafrophone
- Start date

- #1

- 2

- 0

I was trying to change it into a product of an error function and a gamma function, but I needed an extra dx. Any other ideas?

- #2

- 798

- 34

Let X = ? such as : ax+bx² = A X² +B

- #3

- 2

- 0

I tried setting u=ax, v=bx^2. Then i would use

$$ \Gamma(1) = \int_{0}^{\infty} e^{-u}du $$

$${\rm erf}(x) = \int_{-\infty}^{x} e^{-v^2}dv $$

I'm not exactly sure how to use that in

$$ \int_{-\infty}^{+\infty} e^{-u} e^{-v^2} du$$

without another differential dv somewhere.

- #4

- 798

- 34

ax+bx² = b(x+(a/2b))² -a²/4b

X= x+(a/2b)

exp(-(ax+bx²)) = exp(a²/4b)*exp(-b X²)

Then integrate exp(-b X²)*dX

- #5

- 606

- 1

I was trying to change it into a product of an error function and a gamma function, but I needed an extra dx. Any other ideas?

[tex]\displaystyle{ \int_{-\infty}^{+\infty} e^{-(ax+bx^2)} dx= \int_{-\infty}^{+\infty} e^{-b\left(x+\frac{a}{2b}\right)^2}e^{\frac{a^2}{4b}} dx=\frac{e^{\frac{a^2}{4b}}}{\sqrt{b}} \int_{-\infty}^{+\infty} e^{-x^2} dx=e^{\frac{a^2}{4b}}\sqrt{\frac{\pi}{b}}\,,\,\,with}[/tex]

(1) first equality: completing the square

(2) second equality: substituting [itex]\displaystyle{\sqrt{b}\left(x+\frac{a}{2b}\right) \to y}[/itex]

(3) third equality: using [itex]\displaystyle{\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}}[/itex]

DonAntonio

Ps. Of course, I assume [itex]b>0[/itex]

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