Question

For\[\int_{-\infty}^{\infty} e^{tx} \frac{1}{\sigma \sqrt{2\pi}} e^{\frac{-(x-\mu)^2}{2\sigma^2}}dx\]will we get \[e^{t \mu + \frac{1}{2}\sigma^2 t^2}\]?

Answer 1

the small part is hard to see

Answer 2

Hmm power of e?

Answer 3

ya

Answer 4

have u learnt integration by parts yet

Answer 5

\[-\frac{(x-\mu)^2}{2\sigma^2}\]

Answer 6

Yes.

Answer 7

well before you try by parts just expand that bracket
-(x-u)^2 and see if u can simply it down

Answer 8

Hmm.. How are you going to "simplify it"?

Answer 9

like u know that e^(x+a) = same as e^x * e^a

Answer 10

\[\int_{-\infty}^{\infty} e^{tx} \frac{1}{\sigma \sqrt{2\pi}} e^{\frac{-(x-\mu)^2}{2\sigma^2}}dx\]\[=\int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} e^{\frac{-(x-\mu)^2}{2\sigma^2}+tx}dx\]\[=\int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} e^{\frac{-(x-\mu)^2+2\sigma^2tx}{2\sigma^2}}dx\]\[\large =\int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} e^{\frac{-(x^2-2\mu x +\mu^2-2\sigma^2tx)}{2\sigma^2}}dx\]

Answer 11

so