gamma function Query in Schaum’s .

[William Frederick]

Hi , I have hit the wall p76 of Schaums Outlines..Fourier Analysis etc by Murray Spiegel
Q4.22.
Prove that
Integral from 0-infinity of (cos x)/ x^p dx .....................................................[edited]
=
pi / {2 gamma (p) cos (p.pi /2)}
The next step is too quick
Can you please suggest an intermediary step..?:
“We have 1/x^p= {1/gamma(p)} . {the integral from 0- infinity of
[u^(p-1) . e^(-xu) ]du}
Thanks in anticipation ..Bill

 

[HallsofIvy]

That integral is \(\displaystyle \int_0^\infty \frac{cos(x)}{x^p} dx\) isn't it? You forgot the "p" in the exponent. [OP has been edited to correct form.]

 

[William Frederick]

Hi , I have hit the wall p76 of Schaums Outlines..Fourier Analysis etc by Murray Spiegel
Q4.22.
Prove that
Integral from 0-infinity of (cos x)/ x^ p dx
=
pi / {2 gamma (p) cos (p.pi /2)}
The next step is too quick
Can you please suggest an intermediary step..?:
“We have 1/x^p= {1/gamma(p)} . {the integral from 0- infinity of
[u^(p-1) . e^(-xu) ]du}
Thanks in anticipation ..Bill

That integral is \(\displaystyle \int_0^\infty \frac{cos(x)}{x^p} dx\) isn't it? You forgot the "p" in the exponent.

yes i tried to corect that

That integral is \(\displaystyle \int_0^\infty \frac{cos(x)}{x^p} dx\) isn't it? You forgot the "p" in the exponent.

yes how can I edit that ?

 

[Cubist]

I found a scan of the book online, and these are the two lines in question...

Prove that \(\displaystyle \int_0^\infty \frac{cos(x)}{x^p} dx = \frac{\pi}{2 \Gamma (p) cos\left( \frac{p \pi}{2} \right)}, 0<p<1. \)

We have \(\displaystyle \frac{1}{x^p} = \frac{1}{\Gamma (p)} \int_0^\infty u^{p-1} e^{-xu} du \). Then...

--

I think the second line is not a manipulation of the first line. It seems to be derived from the definition of the Gamma function. On the wikipedia page the following is shown as the definition (I changed the variables)...

\(\displaystyle \Gamma (p)=\int_0^\infty v^{p-1}e^{-v} dv \)

If we let v=xu, where x is a constant, then dv = x du

\(\displaystyle \Gamma (p)=\int_0^\infty (xu)^{p-1}e^{-xu} x du \)

\(\displaystyle \Gamma (p)=\int_0^\infty x^{p-1}u^{p-1}e^{-xu} x du \)

\(\displaystyle \Gamma (p)=\int_0^\infty x^p u^{p-1}e^{-xu} du \)

\(\displaystyle \Gamma (p)=x^p \int_0^\infty u^{p-1}e^{-xu} du \)

\(\displaystyle \frac{1}{x^p} = \frac{1}{\Gamma (p)} \int_0^\infty u^{p-1} e^{-xu} du \)

It is likely that this "adjusted standard result" is then used to prove the first line. (I have not looked further). NOTE this is my first foray into the world of the Gamma function, so I may be wrong! (Does the book perhaps derive the above formula on a previous page?)

 

[William Frederick]

yes i tried to corect that

yes how can I edit that ?

I think I have managed to edit it now?thanks

 

[William Frederick]

I found a scan of the book online, and these are the two lines in question...

Prove that \(\displaystyle \int_0^\infty \frac{cos(x)}{x^p} dx = \frac{\pi}{2 \Gamma (p) cos\left( \frac{p \pi}{2} \right)}, 0<p<1. \)

We have \(\displaystyle \frac{1}{x^p} = \frac{1}{\Gamma (p)} \int_0^\infty u^{p-1} e^{-xu} du \). Then...

--

I think the second line is not a manipulation of the first line. It seems to be derived from the definition of the Gamma function. On the wikipedia page the following is shown as the definition (I changed the variables)...

\(\displaystyle \Gamma (p)=\int_0^\infty v^{p-1}e^{-v} dv \)

If we let v=xu, where x is a constant, then dv = x du

\(\displaystyle \Gamma (p)=\int_0^\infty (xu)^{p-1}e^{-xu} x du \)

\(\displaystyle \Gamma (p)=\int_0^\infty x^{p-1}u^{p-1}e^{-xu} x du \)

\(\displaystyle \Gamma (p)=\int_0^\infty x^p u^{p-1}e^{-xu} du \)

\(\displaystyle \Gamma (p)=x^p \int_0^\infty u^{p-1}e^{-xu} du \)

\(\displaystyle \frac{1}{x^p} = \frac{1}{\Gamma (p)} \int_0^\infty u^{p-1} e^{-xu} du \)

It is likely that this "adjusted standard result" is then used to prove the first line. (I have not looked further). NOTE this is my first foray into the world of the Gamma function, so I may be wrong! (Does the book perhaps derive the above formula on a previous page?)

Yes thanks , you are right
I was unable to do the change of variable.. much appreciated.. Bill