Find .

Using , we see

Next, we try to eliminate the *e*^{x} term in the numerator. By noting that , we do so as follows:

.

For the second integral, we make the u-substitution *u*=2x to get

and renaming the variable back to *x*, we get

.

Now, we showed here that . Thus, our integral is

.

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Tags: Gamma Function, Hyperbolic Sine, Integral, Math, Monday Math, Riemann Zeta Function

This entry was posted on June 22, 2009 at 8:11 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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June 29, 2009 at 1:50 am |

[…] parts, with and ; then and Now, using and , integration by parts tells us: . Now, we proved here that , so the above is just this for the case n=2, so since and (see here), then our integral is […]