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...
$$int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi alpha/2)}$$
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 '18 at 4:25
Nosrati
26.5k62353
asked Nov 19 '18 at 3:55
Dhamnekar Winod
385414
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 '18 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 |
$$int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi alpha/2)}$$
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 '18 at 4:25
Nosrati
26.5k62353
asked Nov 19 '18 at 3:55
Dhamnekar Winod
385414
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 '18 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 '18 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 '18 at 4:08
add a comment |
$$int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi alpha/2)}$$
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 '18 at 4:25
Nosrati
26.5k62353
asked Nov 19 '18 at 3:55
Dhamnekar Winod
385414
$$int_0^infty frac{x^{alpha}ln x}{x^2+1},dx=frac{pi^2}{4} frac{sin(pi alpha/2)}{cos^2(pi alpha/2)}$$
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 '18 at 4:25
Nosrati
26.5k62353
asked Nov 19 '18 at 3:55
Dhamnekar Winod
385414
edited Nov 19 '18 at 4:25
Nosrati
26.5k62353
asked Nov 19 '18 at 3:55
Dhamnekar Winod
385414
edited Nov 19 '18 at 4:25
Nosrati
26.5k62353
edited Nov 19 '18 at 4:25
Nosrati
26.5k62353
edited Nov 19 '18 at 4:25
Nosrati
26.5k62353
26.5k62353
asked Nov 19 '18 at 3:55
Dhamnekar Winod
385414
asked Nov 19 '18 at 3:55
Dhamnekar Winod
385414
asked Nov 19 '18 at 3:55
Dhamnekar Winod
385414
385414
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 '18 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 '18 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 '18 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 '18 at 4:08
add a comment |
Have you searched in this site about this integral?
– Nosrati
Nov 19 '18 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 '18 at 4:08
Have you searched in this site about this integral?
– Nosrati
Nov 19 '18 at 3:59
Have you searched in this site about this integral?
– Nosrati
Nov 19 '18 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 '18 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 '18 at 4:08
add a comment |
2 Answers
2
active
oldest
votes
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
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 '18 at 7:25
robjohn♦
264k27303623
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 '18 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 '18 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 '18 at 12:35
add a comment |
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 '18 at 4:15
Nosrati
26.5k62353
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 '18 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 '18 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 '18 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 '18 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 '18 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
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
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 '18 at 7:25
robjohn♦
264k27303623
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 '18 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 '18 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 '18 at 12:35
add a comment |
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
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 '18 at 7:25
robjohn♦
264k27303623
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 '18 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 '18 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 '18 at 12:35
add a comment |
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
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 '18 at 7:25
robjohn♦
264k27303623
$$
begin{align}
int_0^inftyfrac{x^alphalog(x)}{x^2+1},mathrm{d}x
&=frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^alpha}{x^2+1},mathrm{d}xtag1\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}int_0^inftyfrac{x^{frac{alpha-1}2}}{x+1},mathrm{d}xtag2\
&=frac12frac{mathrm{d}}{mathrm{d}alpha}frac{Gammaleft(frac{1+alpha}2right)Gammaleft(frac{1-alpha}2right)}{Gamma(1)}tag3\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}cscleft(pifrac{1-alpha}2right)tag4\
&=fracpi2frac{mathrm{d}}{mathrm{d}alpha}secleft(frac{pialpha}2right)tag5\
&=frac{pi^2}4tanleft(frac{pialpha}2right)secleft(frac{pialpha}2right)tag6
end{align}
$$
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 '18 at 7:25
robjohn♦
264k27303623
answered Nov 19 '18 at 7:25
robjohn♦
264k27303623
answered Nov 19 '18 at 7:25
robjohn♦
264k27303623
answered Nov 19 '18 at 7:25
robjohn♦
264k27303623
264k27303623
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 '18 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 '18 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 '18 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 '18 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 '18 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 '18 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 '18 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 '18 at 13:28
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 '18 at 13:58
What is $mathrm{d}sqrt{x}$? You got the $frac12$, but forgot the $x^{-1/2}$.
– robjohn♦
Nov 19 '18 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 '18 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 '18 at 12:35
add a comment |
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 '18 at 4:15
Nosrati
26.5k62353
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 '18 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 '18 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 '18 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 '18 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 '18 at 6:36
|
show 4 more comments
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 '18 at 4:15
Nosrati
26.5k62353
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 '18 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 '18 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 '18 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 '18 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 '18 at 6:36
|
show 4 more comments
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 '18 at 4:15
Nosrati
26.5k62353
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 '18 at 4:15
Nosrati
26.5k62353
edited Nov 19 '18 at 5:33
answered Nov 19 '18 at 4:15
Nosrati
26.5k62353
answered Nov 19 '18 at 4:15
Nosrati
26.5k62353
answered Nov 19 '18 at 4:15
Nosrati
26.5k62353
26.5k62353
1
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 '18 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 '18 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 '18 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 '18 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 '18 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 '18 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 '18 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 '18 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 '18 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 '18 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 '18 at 4:59
How did you compute the first equation? I never found it in the list of trigonometric identities?
– Dhamnekar Winod
Nov 19 '18 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 '18 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 '18 at 10:48
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 '18 at 5:35
@Nosrati, is $I(a)$ regularised incomplete beta function? why it is multiplied by$frac12$?
– Dhamnekar Winod
Nov 20 '18 at 5:35
See the link, fourth formula of Properties.
– Nosrati
Nov 20 '18 at 5:37
See the link, fourth formula of Properties.
– Nosrati
Nov 20 '18 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 '18 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 '18 at 6:36
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Have you searched in this site about this integral?
– Nosrati
Nov 19 '18 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 '18 at 4:08