Question

I have an integral of (a/(t+a))^b. With substitution I can get to (a^b * u^(1-b))/(1-b) - that is [numerator/(1-b)] however, when I substitute the (t+a) for u back into the thang I supposed to somehow (magically) negate the result -essentially getting [-numerator/(b-1)]. What is the "black magic" I am missing? Cheers, A

Answer 1

\[\int\limits_{0}^{\infty} \left( a / t+a \right)^{b}\]

Answer 2

\[a ^{b} \int\limits\limits_{0}^{\infty} \left( \frac{ 1 }{ t+a } \right)^{b} dt\]

Answer 3

\[a ^{b}\int\limits_{0}^{\infty} \left( t+a \right)^{-b} dt\]

Answer 4

u=t+a, du=dt etc.

Answer 5

\[\frac{ a ^{b}u ^{1-b} }{1-b }\]

Answer 6

So I get to the equation above but I am vague on how after substituting u=t+a that you then get \[-\frac{ a ^{b}\left( a+t \right)^{1-b} }{b-1 }\]

Answer 7

If it helps, the worked example is attached.

Answer 8

what's the issue?

you should have a limit somwhere...
\[\lim_{t \rightarrow infnty}F(t) - F(0)\text{ where }F(t) = \text {the Integral once integrated}\]

Answer 9

My problem, I am trying to work out the magic in the answer provided (see attached).

\[\int\limits\limits_{0}^{\infty} \left( a / t+a \right)^{b} dt\]

Is processed in background into:

\[\left[ -\frac{ a ^{b}\left( a+t \right)^{1-b} }{b-1 } \right]\]

When trying to get to that point myself (while trying to remember my calculus from 20+ years ago) I get to, with u-substitution:

\[\frac{ a ^{b}u ^{1-b} }{1-b }\]

The question is what steps, rules, assumptions am I missing that would get me from:
\[\frac{ a ^{b}u ^{1-b} }{1-b }\] to \[\left[ -\frac{ a ^{b}\left( a+t \right)^{1-b} }{b-1 } \right]\]

Answer 10

you have to evaluate...\[\lim_{t \rightarrow \infty}\left[ \frac{ a^b(t+a)^{1-b}}{1-b} \right]-\frac{ a^b((0)+a)^{1-b}}{1-b}\]

Answer 11

Message repeats, what steps, rules, assumptions am I missing that would get me from:
\[\frac{ a ^{b}u ^{1-b} }{1-b }\]
to

\[\[\left[ -\frac{ a ^{b}\left( a+t \right)^{1-b} }{b-1 } \right]\]\]

Why has substituting back for u=a+t negated the function and reversed the factors in the denominator?

I haven't been in a class room for over 30 years and am trying to relearn quite a lot for my Masters so pardon my simple mind ;) It may be quite a basic high school level trick that has been lost to me after all those years of red wine ;)

Answer 12

you have to evaluate at the limits of integration.

Answer 13

Yep that is the goal.

Answer 14

I note you have the 1-b in the denominator but the solution I am trying to reverse engineer as b-1. My confusion since my math is so rusty is how the 1-b becomes b-1.

Answer 15

\[\int_{0}^{\infty}\left( \frac{ a }{ t+a } \right)^b dt=\left.\frac{ a^b(t+a)^{1-b} }{1-b } \right]^{\infty}_0= \lim_{t \rightarrow \infty}\left[ \frac{ a^b(t+a)^{1-b} }{1-b } \right] - \frac{ a^b(0+a)^{1-b} }{1-b }\]

Answer 16

so b>1 in order for the first term to go to 0 as t goes to infinity.

Answer 17

the second term gives the result.

Answer 18

I've attached the solution I am trying to reverse engineer. I am okay with the notion that one term goes to 0 and I guess was getting to the notion that the aspect at "2" was actually short hand for what you have scratched out for me above.

Answer 19

so beta > 1. 1/infinity -> 0

it's all there.

Answer 20

However, notice that in the its (b-1) where as you have (1-b), now I have got through u-substitution (1-b). I note using tools on the net the tools swap from (1-b) to (b-1) once u=t+a is re-substituted back in.

Answer 21

That's my problem I guess, I was moving to the solution you came to. Checked the provided answer, noted the b and 1 swapped in denominator, checked wolfram website and it had (1-b) up to expanding u again after which it became (b-1).

Answer 22

they messed up on step 2.

Answer 23

Maybe, but attached is the wolfram sites go at it so I agree the exponent b-1 in the numerator might be wrong but there is that pesky b-1 in the denominator. A couple of other web tools tend to do the same thing.

Answer 24

Thanks for your help.