Mixing
General formula for $int_0^{pi/2} tan^{alpha}(x) dx$?
$begingroup$
There are already questions about how to find $int tan^{1/2}(x) dx$. But how to derive a general formula for $int_0^{pi/2} tan^{alpha}(x) dx$ (which converges if $|alpha|<1$) ?
More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=tan^2(x)$ in $int_0^infty frac{v^beta}{1+v} dv$
calculus integration improper-integrals gamma-function
asked Feb 1 at 3:39
Holding ArthurHolding Arthur
1,555417
$endgroup$
add a comment |
$begingroup$
There are already questions about how to find $int tan^{1/2}(x) dx$. But how to derive a general formula for $int_0^{pi/2} tan^{alpha}(x) dx$ (which converges if $|alpha|<1$) ?
More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=tan^2(x)$ in $int_0^infty frac{v^beta}{1+v} dv$
calculus integration improper-integrals gamma-function
asked Feb 1 at 3:39
Holding ArthurHolding Arthur
1,555417
$endgroup$
$begingroup$
Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
$endgroup$
– robjohn♦
Feb 1 at 4:03
add a comment |
$begingroup$
There are already questions about how to find $int tan^{1/2}(x) dx$. But how to derive a general formula for $int_0^{pi/2} tan^{alpha}(x) dx$ (which converges if $|alpha|<1$) ?
More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=tan^2(x)$ in $int_0^infty frac{v^beta}{1+v} dv$
calculus integration improper-integrals gamma-function
asked Feb 1 at 3:39
Holding ArthurHolding Arthur
1,555417
$endgroup$
There are already questions about how to find $int tan^{1/2}(x) dx$. But how to derive a general formula for $int_0^{pi/2} tan^{alpha}(x) dx$ (which converges if $|alpha|<1$) ?
More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=tan^2(x)$ in $int_0^infty frac{v^beta}{1+v} dv$
calculus integration improper-integrals gamma-function
calculus integration improper-integrals gamma-function
asked Feb 1 at 3:39
Holding ArthurHolding Arthur
1,555417
asked Feb 1 at 3:39
Holding ArthurHolding Arthur
1,555417
asked Feb 1 at 3:39
Holding ArthurHolding Arthur
1,555417
asked Feb 1 at 3:39
Holding ArthurHolding Arthur
1,555417
asked Feb 1 at 3:39
Holding ArthurHolding Arthur
1,555417
1,555417
$begingroup$
Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
$endgroup$
– robjohn♦
Feb 1 at 4:03
add a comment |
$begingroup$
Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
$endgroup$
– robjohn♦
Feb 1 at 4:03
$begingroup$
Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
$endgroup$
– robjohn♦
Feb 1 at 4:03
$begingroup$
Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
$endgroup$
– robjohn♦
Feb 1 at 4:03
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
With $u=tan x$, we have
$$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
Now, according to this question:
$$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
So, plugging in $beta=frac pi 2$ yields
$$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$
answered Feb 1 at 3:56
Stefan LafonStefan Lafon
3,005212
$endgroup$
add a comment |
$begingroup$
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
Setting $t=sin(x)^2$:
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
Then recall the definition of the beta function
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
We use this to see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
Hence
$$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
It can be easily shown that
$$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
So
$$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
Which you know works for $|a|<1$.
This can be used to show that
$$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
Or simply
$$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.
answered Feb 1 at 4:25
clathratusclathratus
5,1141439
$endgroup$
add a comment |
$begingroup$
$$
begin{align}
int_0^{pi/2}tan^alpha(x),mathrm{d}x
&=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
&=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
&=fracpi2secleft(frac{pialpha}2right)tag3
end{align}
$$
Explanation:
$(1)$: substitute $xmapstoarctan(x)$
$(2)$: use this answer
$(3)$: trigonometric identity
answered Feb 1 at 3:57
robjohn♦robjohn
270k27313642
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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active
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active
oldest
votes
$begingroup$
With $u=tan x$, we have
$$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
Now, according to this question:
$$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
So, plugging in $beta=frac pi 2$ yields
$$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$
answered Feb 1 at 3:56
Stefan LafonStefan Lafon
3,005212
$endgroup$
add a comment |
$begingroup$
With $u=tan x$, we have
$$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
Now, according to this question:
$$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
So, plugging in $beta=frac pi 2$ yields
$$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$
answered Feb 1 at 3:56
Stefan LafonStefan Lafon
3,005212
$endgroup$
add a comment |
$begingroup$
With $u=tan x$, we have
$$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
Now, according to this question:
$$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
So, plugging in $beta=frac pi 2$ yields
$$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$
answered Feb 1 at 3:56
Stefan LafonStefan Lafon
3,005212
$endgroup$
With $u=tan x$, we have
$$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
Now, according to this question:
$$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
So, plugging in $beta=frac pi 2$ yields
$$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$
answered Feb 1 at 3:56
Stefan LafonStefan Lafon
3,005212
answered Feb 1 at 3:56
Stefan LafonStefan Lafon
3,005212
answered Feb 1 at 3:56
Stefan LafonStefan Lafon
3,005212
answered Feb 1 at 3:56
Stefan LafonStefan Lafon
3,005212
3,005212
add a comment |
add a comment |
$begingroup$
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
Setting $t=sin(x)^2$:
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
Then recall the definition of the beta function
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
We use this to see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
Hence
$$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
It can be easily shown that
$$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
So
$$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
Which you know works for $|a|<1$.
This can be used to show that
$$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
Or simply
$$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.
answered Feb 1 at 4:25
clathratusclathratus
5,1141439
$endgroup$
add a comment |
$begingroup$
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
Setting $t=sin(x)^2$:
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
Then recall the definition of the beta function
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
We use this to see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
Hence
$$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
It can be easily shown that
$$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
So
$$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
Which you know works for $|a|<1$.
This can be used to show that
$$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
Or simply
$$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.
answered Feb 1 at 4:25
clathratusclathratus
5,1141439
$endgroup$
add a comment |
$begingroup$
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
Setting $t=sin(x)^2$:
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
Then recall the definition of the beta function
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
We use this to see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
Hence
$$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
It can be easily shown that
$$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
So
$$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
Which you know works for $|a|<1$.
This can be used to show that
$$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
Or simply
$$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.
answered Feb 1 at 4:25
clathratusclathratus
5,1141439
$endgroup$
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
Setting $t=sin(x)^2$:
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
Then recall the definition of the beta function
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
We use this to see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
Hence
$$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
It can be easily shown that
$$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
So
$$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
Which you know works for $|a|<1$.
This can be used to show that
$$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
Or simply
$$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.
answered Feb 1 at 4:25
clathratusclathratus
5,1141439
edited Feb 1 at 4:32
answered Feb 1 at 4:25
clathratusclathratus
5,1141439
answered Feb 1 at 4:25
clathratusclathratus
5,1141439
answered Feb 1 at 4:25
clathratusclathratus
5,1141439
5,1141439
add a comment |
add a comment |
$begingroup$
$$
begin{align}
int_0^{pi/2}tan^alpha(x),mathrm{d}x
&=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
&=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
&=fracpi2secleft(frac{pialpha}2right)tag3
end{align}
$$
Explanation:
$(1)$: substitute $xmapstoarctan(x)$
$(2)$: use this answer
$(3)$: trigonometric identity
answered Feb 1 at 3:57
robjohn♦robjohn
270k27313642
$endgroup$
add a comment |
$begingroup$
$$
begin{align}
int_0^{pi/2}tan^alpha(x),mathrm{d}x
&=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
&=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
&=fracpi2secleft(frac{pialpha}2right)tag3
end{align}
$$
Explanation:
$(1)$: substitute $xmapstoarctan(x)$
$(2)$: use this answer
$(3)$: trigonometric identity
answered Feb 1 at 3:57
robjohn♦robjohn
270k27313642
$endgroup$
add a comment |
$begingroup$
$$
begin{align}
int_0^{pi/2}tan^alpha(x),mathrm{d}x
&=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
&=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
&=fracpi2secleft(frac{pialpha}2right)tag3
end{align}
$$
Explanation:
$(1)$: substitute $xmapstoarctan(x)$
$(2)$: use this answer
$(3)$: trigonometric identity
answered Feb 1 at 3:57
robjohn♦robjohn
270k27313642
$endgroup$
$$
begin{align}
int_0^{pi/2}tan^alpha(x),mathrm{d}x
&=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
&=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
&=fracpi2secleft(frac{pialpha}2right)tag3
end{align}
$$
Explanation:
$(1)$: substitute $xmapstoarctan(x)$
$(2)$: use this answer
$(3)$: trigonometric identity
answered Feb 1 at 3:57
robjohn♦robjohn
270k27313642
edited Feb 1 at 4:56
answered Feb 1 at 3:57
robjohn♦robjohn
270k27313642
answered Feb 1 at 3:57
robjohn♦robjohn
270k27313642
answered Feb 1 at 3:57
robjohn♦robjohn
270k27313642
270k27313642
add a comment |
add a comment |
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$begingroup$
Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
$endgroup$
– robjohn♦
Feb 1 at 4:03