Improper integral $int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi...
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$$int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi alpha/2)}$$
where $0 < alpha < 1$.
Answer: When i put this term in my integral calculator, it gave me very lengthy answer involving polylogarithm functions.
calculus integration improper-integrals self-learning polylogarithm
edited Nov 19 at 4:25
Nosrati
26.3k62353
asked Nov 19 at 3:55
Dhamnekar Winod
361414
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 at 11:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-3
down vote
favorite
$$int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi alpha/2)}$$
where $0 < alpha < 1$.
Answer: When i put this term in my integral calculator, it gave me very lengthy answer involving polylogarithm functions.
calculus integration improper-integrals self-learning polylogarithm
edited Nov 19 at 4:25
Nosrati
26.3k62353
asked Nov 19 at 3:55
Dhamnekar Winod
361414
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 at 11:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
Have you searched in this site about this integral?
– Nosrati
Nov 19 at 3:59
@Nosrati, I didn't search in this site about this integral. But if this is a duplicate question, please provide the link of this question's answer available in this site.
– Dhamnekar Winod
Nov 19 at 4:08
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
$$int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi alpha/2)}$$
where $0 < alpha < 1$.
Answer: When i put this term in my integral calculator, it gave me very lengthy answer involving polylogarithm functions.
calculus integration improper-integrals self-learning polylogarithm
edited Nov 19 at 4:25
Nosrati
26.3k62353
asked Nov 19 at 3:55
Dhamnekar Winod
361414
$$int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi alpha/2)}$$
where $0 < alpha < 1$.
Answer: When i put this term in my integral calculator, it gave me very lengthy answer involving polylogarithm functions.
calculus integration improper-integrals self-learning polylogarithm
calculus integration improper-integrals self-learning polylogarithm
edited Nov 19 at 4:25
Nosrati
26.3k62353
asked Nov 19 at 3:55
Dhamnekar Winod
361414
edited Nov 19 at 4:25
Nosrati
26.3k62353
asked Nov 19 at 3:55
Dhamnekar Winod
361414
edited Nov 19 at 4:25
Nosrati
26.3k62353
edited Nov 19 at 4:25
Nosrati
26.3k62353
edited Nov 19 at 4:25
Nosrati
26.3k62353
26.3k62353
asked Nov 19 at 3:55
Dhamnekar Winod
361414
asked Nov 19 at 3:55
Dhamnekar Winod
361414
asked Nov 19 at 3:55
Dhamnekar Winod
361414
361414
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 at 11:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 at 11:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
Have you searched in this site about this integral?
– Nosrati
Nov 19 at 3:59
@Nosrati, I didn't search in this site about this integral. But if this is a duplicate question, please provide the link of this question's answer available in this site.
– Dhamnekar Winod
Nov 19 at 4:08
add a comment |
Have you searched in this site about this integral?
– Nosrati
Nov 19 at 3:59
@Nosrati, I didn't search in this site about this integral. But if this is a duplicate question, please provide the link of this question's answer available in this site.
– Dhamnekar Winod
Nov 19 at 4:08
Have you searched in this site about this integral?
– Nosrati
Nov 19 at 3:59
Have you searched in this site about this integral?
– Nosrati
Nov 19 at 3:59
@Nosrati, I didn't search in this site about this integral. But if this is a duplicate question, please provide the link of this question's answer available in this site.
– Dhamnekar Winod
Nov 19 at 4:08
@Nosrati, I didn't search in this site about this integral. But if this is a duplicate question, please provide the link of this question's answer available in this site.
– Dhamnekar Winod
Nov 19 at 4:08
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
Explanation:
$(1)$: $frac{mathrm{d}}{mathrm{d}alpha}x^alpha=x^alphalog(x)$
$(2)$: substitute $xmapstosqrt{x}$
$(3)$: Beta Function
$(4)$: Euler Reflection Formula
$(5)$: trigonometric identity
$(6)$: evaluate the derivative
answered Nov 19 at 7:25
robjohn♦
263k27301623
How did you compute (2)? You substitute x by $sqrt{x}$.So I get $frac12frac{d}{dalpha}intlimits_0^inftyfrac{x^frac{alpha}{2}}{x+1}dx$ But your term in (2) is different?
– Dhamnekar Winod
Nov 19 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 at 13:58
I already got the answers to this question from you and one other member of this site correctly satisfying all my queries. Then what is the use of making it off-topic and put on hold?
– Dhamnekar Winod
Nov 25 at 12:35
add a comment |
up vote
4
down vote
By substitution $x=tan t$
$$I(a)=int_0^inftydfrac{x^a}{1+x^2} dx=int_0^{pi/2}tan^at dt$$
then using dear Beta function
$$I(a)=dfrac12Gammaleft(dfrac{a+1}{2}right)Gammaleft(dfrac{-a+1}{2}right)=dfrac{pi}{2sinpidfrac{1-a}{2}}=dfrac{pi}{2cosdfrac{api}{2}}$$
by Reflection formula the desired integral is
$$dfrac{d}{da}I(a)=color{blue}{dfrac{pi^2}{4}dfrac{sinfrac{api}{2}}{cos^2frac{api}{2}}}$$
answered Nov 19 at 4:15
Nosrati
26.3k62353
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 at 4:59
1
It is in fact no trigonometric identity but just the substitution $x=tan(t)$. From hereon you will get the upper as well as the lower border of integration. Further note that the derivative of $x=tan(t)$ can be written as $dt=frac{dx}{1+x^2}$ due the derivative of the tangent function.
– mrtaurho
Nov 19 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 at 5:37
1
So $$I(a)=int_0^{pi/2}tan^at dt=dfrac12left(2int_0^{pi/2}sin^atcos^{-a}t dtright)$$
– Nosrati
Nov 20 at 6:36
|
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
Explanation:
$(1)$: $frac{mathrm{d}}{mathrm{d}alpha}x^alpha=x^alphalog(x)$
$(2)$: substitute $xmapstosqrt{x}$
$(3)$: Beta Function
$(4)$: Euler Reflection Formula
$(5)$: trigonometric identity
$(6)$: evaluate the derivative
answered Nov 19 at 7:25
robjohn♦
263k27301623
How did you compute (2)? You substitute x by $sqrt{x}$.So I get $frac12frac{d}{dalpha}intlimits_0^inftyfrac{x^frac{alpha}{2}}{x+1}dx$ But your term in (2) is different?
– Dhamnekar Winod
Nov 19 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 at 13:58
I already got the answers to this question from you and one other member of this site correctly satisfying all my queries. Then what is the use of making it off-topic and put on hold?
– Dhamnekar Winod
Nov 25 at 12:35
add a comment |
up vote
5
down vote
accepted
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
Explanation:
$(1)$: $frac{mathrm{d}}{mathrm{d}alpha}x^alpha=x^alphalog(x)$
$(2)$: substitute $xmapstosqrt{x}$
$(3)$: Beta Function
$(4)$: Euler Reflection Formula
$(5)$: trigonometric identity
$(6)$: evaluate the derivative
answered Nov 19 at 7:25
robjohn♦
263k27301623
How did you compute (2)? You substitute x by $sqrt{x}$.So I get $frac12frac{d}{dalpha}intlimits_0^inftyfrac{x^frac{alpha}{2}}{x+1}dx$ But your term in (2) is different?
– Dhamnekar Winod
Nov 19 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 at 13:58
I already got the answers to this question from you and one other member of this site correctly satisfying all my queries. Then what is the use of making it off-topic and put on hold?
– Dhamnekar Winod
Nov 25 at 12:35
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
Explanation:
$(1)$: $frac{mathrm{d}}{mathrm{d}alpha}x^alpha=x^alphalog(x)$
$(2)$: substitute $xmapstosqrt{x}$
$(3)$: Beta Function
$(4)$: Euler Reflection Formula
$(5)$: trigonometric identity
$(6)$: evaluate the derivative
answered Nov 19 at 7:25
robjohn♦
263k27301623
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
Explanation:
$(1)$: $frac{mathrm{d}}{mathrm{d}alpha}x^alpha=x^alphalog(x)$
$(2)$: substitute $xmapstosqrt{x}$
$(3)$: Beta Function
$(4)$: Euler Reflection Formula
$(5)$: trigonometric identity
$(6)$: evaluate the derivative
answered Nov 19 at 7:25
robjohn♦
263k27301623
answered Nov 19 at 7:25
robjohn♦
263k27301623
answered Nov 19 at 7:25
robjohn♦
263k27301623
answered Nov 19 at 7:25
robjohn♦
263k27301623
263k27301623
How did you compute (2)? You substitute x by $sqrt{x}$.So I get $frac12frac{d}{dalpha}intlimits_0^inftyfrac{x^frac{alpha}{2}}{x+1}dx$ But your term in (2) is different?
– Dhamnekar Winod
Nov 19 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 at 13:58
I already got the answers to this question from you and one other member of this site correctly satisfying all my queries. Then what is the use of making it off-topic and put on hold?
– Dhamnekar Winod
Nov 25 at 12:35
add a comment |
How did you compute (2)? You substitute x by $sqrt{x}$.So I get $frac12frac{d}{dalpha}intlimits_0^inftyfrac{x^frac{alpha}{2}}{x+1}dx$ But your term in (2) is different?
– Dhamnekar Winod
Nov 19 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 at 13:58
I already got the answers to this question from you and one other member of this site correctly satisfying all my queries. Then what is the use of making it off-topic and put on hold?
– Dhamnekar Winod
Nov 25 at 12:35
How did you compute (2)? You substitute x by $sqrt{x}$.So I get $frac12frac{d}{dalpha}intlimits_0^inftyfrac{x^frac{alpha}{2}}{x+1}dx$ But your term in (2) is different?
– Dhamnekar Winod
Nov 19 at 13:28
How did you compute (2)? You substitute x by $sqrt{x}$.So I get $frac12frac{d}{dalpha}intlimits_0^inftyfrac{x^frac{alpha}{2}}{x+1}dx$ But your term in (2) is different?
– Dhamnekar Winod
Nov 19 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 at 13:58
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 at 13:58
I already got the answers to this question from you and one other member of this site correctly satisfying all my queries. Then what is the use of making it off-topic and put on hold?
– Dhamnekar Winod
Nov 25 at 12:35
I already got the answers to this question from you and one other member of this site correctly satisfying all my queries. Then what is the use of making it off-topic and put on hold?
– Dhamnekar Winod
Nov 25 at 12:35
add a comment |
up vote
4
down vote
By substitution $x=tan t$
$$I(a)=int_0^inftydfrac{x^a}{1+x^2} dx=int_0^{pi/2}tan^at dt$$
then using dear Beta function
$$I(a)=dfrac12Gammaleft(dfrac{a+1}{2}right)Gammaleft(dfrac{-a+1}{2}right)=dfrac{pi}{2sinpidfrac{1-a}{2}}=dfrac{pi}{2cosdfrac{api}{2}}$$
by Reflection formula the desired integral is
$$dfrac{d}{da}I(a)=color{blue}{dfrac{pi^2}{4}dfrac{sinfrac{api}{2}}{cos^2frac{api}{2}}}$$
answered Nov 19 at 4:15
Nosrati
26.3k62353
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 at 4:59
1
It is in fact no trigonometric identity but just the substitution $x=tan(t)$. From hereon you will get the upper as well as the lower border of integration. Further note that the derivative of $x=tan(t)$ can be written as $dt=frac{dx}{1+x^2}$ due the derivative of the tangent function.
– mrtaurho
Nov 19 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 at 5:37
1
So $$I(a)=int_0^{pi/2}tan^at dt=dfrac12left(2int_0^{pi/2}sin^atcos^{-a}t dtright)$$
– Nosrati
Nov 20 at 6:36
|
show 4 more comments
up vote
4
down vote
By substitution $x=tan t$
$$I(a)=int_0^inftydfrac{x^a}{1+x^2} dx=int_0^{pi/2}tan^at dt$$
then using dear Beta function
$$I(a)=dfrac12Gammaleft(dfrac{a+1}{2}right)Gammaleft(dfrac{-a+1}{2}right)=dfrac{pi}{2sinpidfrac{1-a}{2}}=dfrac{pi}{2cosdfrac{api}{2}}$$
by Reflection formula the desired integral is
$$dfrac{d}{da}I(a)=color{blue}{dfrac{pi^2}{4}dfrac{sinfrac{api}{2}}{cos^2frac{api}{2}}}$$
answered Nov 19 at 4:15
Nosrati
26.3k62353
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 at 4:59
1
It is in fact no trigonometric identity but just the substitution $x=tan(t)$. From hereon you will get the upper as well as the lower border of integration. Further note that the derivative of $x=tan(t)$ can be written as $dt=frac{dx}{1+x^2}$ due the derivative of the tangent function.
– mrtaurho
Nov 19 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 at 5:37
1
So $$I(a)=int_0^{pi/2}tan^at dt=dfrac12left(2int_0^{pi/2}sin^atcos^{-a}t dtright)$$
– Nosrati
Nov 20 at 6:36
|
show 4 more comments
up vote
4
down vote
up vote
4
down vote
By substitution $x=tan t$
$$I(a)=int_0^inftydfrac{x^a}{1+x^2} dx=int_0^{pi/2}tan^at dt$$
then using dear Beta function
$$I(a)=dfrac12Gammaleft(dfrac{a+1}{2}right)Gammaleft(dfrac{-a+1}{2}right)=dfrac{pi}{2sinpidfrac{1-a}{2}}=dfrac{pi}{2cosdfrac{api}{2}}$$
by Reflection formula the desired integral is
$$dfrac{d}{da}I(a)=color{blue}{dfrac{pi^2}{4}dfrac{sinfrac{api}{2}}{cos^2frac{api}{2}}}$$
answered Nov 19 at 4:15
Nosrati
26.3k62353
By substitution $x=tan t$
$$I(a)=int_0^inftydfrac{x^a}{1+x^2} dx=int_0^{pi/2}tan^at dt$$
then using dear Beta function
$$I(a)=dfrac12Gammaleft(dfrac{a+1}{2}right)Gammaleft(dfrac{-a+1}{2}right)=dfrac{pi}{2sinpidfrac{1-a}{2}}=dfrac{pi}{2cosdfrac{api}{2}}$$
by Reflection formula the desired integral is
$$dfrac{d}{da}I(a)=color{blue}{dfrac{pi^2}{4}dfrac{sinfrac{api}{2}}{cos^2frac{api}{2}}}$$
answered Nov 19 at 4:15
Nosrati
26.3k62353
edited Nov 19 at 5:33
answered Nov 19 at 4:15
Nosrati
26.3k62353
answered Nov 19 at 4:15
Nosrati
26.3k62353
answered Nov 19 at 4:15
Nosrati
26.3k62353
26.3k62353
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 at 4:59
1
It is in fact no trigonometric identity but just the substitution $x=tan(t)$. From hereon you will get the upper as well as the lower border of integration. Further note that the derivative of $x=tan(t)$ can be written as $dt=frac{dx}{1+x^2}$ due the derivative of the tangent function.
– mrtaurho
Nov 19 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 at 5:37
1
So $$I(a)=int_0^{pi/2}tan^at dt=dfrac12left(2int_0^{pi/2}sin^atcos^{-a}t dtright)$$
– Nosrati
Nov 20 at 6:36
|
show 4 more comments
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 at 4:59
1
It is in fact no trigonometric identity but just the substitution $x=tan(t)$. From hereon you will get the upper as well as the lower border of integration. Further note that the derivative of $x=tan(t)$ can be written as $dt=frac{dx}{1+x^2}$ due the derivative of the tangent function.
– mrtaurho
Nov 19 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 at 5:37
1
So $$I(a)=int_0^{pi/2}tan^at dt=dfrac12left(2int_0^{pi/2}sin^atcos^{-a}t dtright)$$
– Nosrati
Nov 20 at 6:36
1
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 at 4:59
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 at 4:59
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It is in fact no trigonometric identity but just the substitution $x=tan(t)$. From hereon you will get the upper as well as the lower border of integration. Further note that the derivative of $x=tan(t)$ can be written as $dt=frac{dx}{1+x^2}$ due the derivative of the tangent function.
– mrtaurho
Nov 19 at 10:48
It is in fact no trigonometric identity but just the substitution $x=tan(t)$. From hereon you will get the upper as well as the lower border of integration. Further note that the derivative of $x=tan(t)$ can be written as $dt=frac{dx}{1+x^2}$ due the derivative of the tangent function.
– mrtaurho
Nov 19 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 at 5:35
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 at 5:37
See the link, fourth formula of Properties.
– Nosrati
Nov 20 at 5:37
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So $$I(a)=int_0^{pi/2}tan^at dt=dfrac12left(2int_0^{pi/2}sin^atcos^{-a}t dtright)$$
– Nosrati
Nov 20 at 6:36
So $$I(a)=int_0^{pi/2}tan^at dt=dfrac12left(2int_0^{pi/2}sin^atcos^{-a}t dtright)$$
– Nosrati
Nov 20 at 6:36
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Have you searched in this site about this integral?
– Nosrati
Nov 19 at 3:59
@Nosrati, I didn't search in this site about this integral. But if this is a duplicate question, please provide the link of this question's answer available in this site.
– Dhamnekar Winod
Nov 19 at 4:08