Integration by Parts?

**diddledabble** · July 22nd 2009, 09:31 AM

$ \int_0^1 (1-t) e^{-st} dt $

[tex]
\int_0^1 e^{2t-st} dt
[\math]

Would the be done by parts method? I am stuck please help.

**e^(i*pi)** · July 22nd 2009, 09:39 AM

Originally Posted by diddledabble

$ \int_0^1 (1-t) e^{-st} dt $
[tex]
\int_0^{\infty} e^{(2-s)t} dt
[\math]
Would the be done by parts method? I am stuck please help.

$u = 1-t$
$\frac{du}{dt} = -1$

$\frac{dv}{dt} = e^{st}$
$v = \frac{1}{s}e^{st}$

That -1 should mean only one iteration is needed

**DeMath** · July 22nd 2009, 09:49 AM

Originally Posted by diddledabble

$ \int_0^1 (1-t) e^{-st} dt$

$\int_0^1 e^{2t-st} dt$

Would the be done by parts method? I am stuck please help.

$\int\limits_0^1 {\left( {1 - t} \right){e^{ - st}}dt} = - \frac{1} {s}\int\limits_0^1 {\left( {1 - t} \right)d\left( {{e^{ - st}}} \right)} = \left. { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right)} \right|_{t = 0}^{t = 1}$ $+ \frac{1}{s}\int\limits_0^1 {{e^{ - st}}d\left( {1 - t} \right)} =$

$= \left. { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right)} \right|_{t = 0}^{t = 1} - \frac{1}{s}\int\limits_0^1 {{e^{ - st}}dt} = \left. { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right)} \right|_{t = 0}^{t = 1} + \frac{1} {{{s^2}}}\int\limits_0^1 {d\left( {{e^{ - st}}} \right)} =$

$= \left. {\left( { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right) + \frac{{{e^{ - st}}}}{{{s^2}}}} \right)} \right|_{t = 0}^{t = 1} = \frac{{{e^{ - s}}}}{{{s^2}}} + \frac{1}{s} - \frac{1}{{{s^2}}} = \frac{1} {{{s^2}}}\left( {{e^{ - s}} + s - 1} \right).$

**diddledabble** · July 22nd 2009, 09:55 AM

I tried typing this one in Latex but it wouldn't come out.

[tex]
\int_0^1 e^{(2-s)t} dt
[\math]

**pomp** · July 22nd 2009, 10:13 AM

Originally Posted by diddledabble

I tried typing this one in Latex but it wouldn't come out.

[tex]
\int_0^1 e^{(2-s)t} dt
[\math]

/math not \math

**diddledabble** · July 22nd 2009, 10:22 AM

$ \int_0^1 e^{(2-s)t} dt $

**DeMath** · July 22nd 2009, 10:31 AM

Originally Posted by diddledabble

$ \int_0^1 e^{(2-s)t} dt $

It's not by parts

$\int\limits_0^1 {{e^{(2 - s)t}}dt} = \frac{1}{{2 - s}}\int\limits_0^1 {d\left( {{e^{(2 - s)t}}} \right)} = \left. {\frac{{{e^{(2 - s)t}}}}{{2 - s}}} \right|_{t = 0}^{t = 1} = \frac{{{e^{2 - s}} - 1}}{{2 - s}}.$

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