Integral $intlimits_0^y text{d}x sin(a x+b/x)/x$Evaluating $intsqrt{150^2-x^2} cdot dx$Evaluate the integral...
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Integral $intlimits_0^y text{d}x sin(a x+b/x)/x$
Evaluating $intsqrt{150^2-x^2} cdot dx$Evaluate the integral $I=intlimits_{-infty}^{infty} frac{sin^{2}u}{u^{2}}du$Compute $intfrac{1-x}{(x-2)(x+3)}$ and $intfrac{cos(3x)}{sin(3x)}$Integral $int{frac{x^5}{sqrt{x^2+7}}},text{d}x$.How to evaluate $I=intlimits_0^{pi/2}frac{xlog{sin{(x)}}}{sin(x)},dx$Integrals with the special functions $Ci(x)$ and $erf(x)$Yet another log-sin integral $intlimits_0^{pi/3}log(1+sin x)log(1-sin x),dx$Calculating $intlimits_0^infty sumlimits_{n=1}^infty f_n(x) ; dx$$intlimits^infty_{-infty} xe^{-|(x-u)|} dx = ?$Integral $intlimits_0^infty text{d}q cos(qx) (omega^2-q^2)^{N-1} e^{-s(omega^2-q^2)^N}$
$begingroup$
I am interested in this integral (for $a,binmathbb{R}$):
$$ F(a,b;y) := intlimits_0^y text{d}x , frac{sin(ax+b/x)}{x} $$
Usually for integrals of this type it is useful to transform $tilde{x} = 1/x$, but in my case the integral boundaries will get changed as well:
$$ F(a,b;y) := intlimits^infty_{1/y} text{d}tilde{x} , frac{sin(btilde{x}+a/tilde{x})}{tilde{x}} $$
So I conclude that this transformation is only useful for the integral over the entire real line, and not in my case.
In the case of $b=0$ one has of course $F(a,0;y) = text{Si}(a,y)$. Is there a special function for the case $bnot=0$? Any ideas?
calculus integration trigonometry definite-integrals
edited yesterday
rash
20811
asked yesterday
JensJens
458
$endgroup$
add a comment |
$begingroup$
I am interested in this integral (for $a,binmathbb{R}$):
$$ F(a,b;y) := intlimits_0^y text{d}x , frac{sin(ax+b/x)}{x} $$
Usually for integrals of this type it is useful to transform $tilde{x} = 1/x$, but in my case the integral boundaries will get changed as well:
$$ F(a,b;y) := intlimits^infty_{1/y} text{d}tilde{x} , frac{sin(btilde{x}+a/tilde{x})}{tilde{x}} $$
So I conclude that this transformation is only useful for the integral over the entire real line, and not in my case.
In the case of $b=0$ one has of course $F(a,0;y) = text{Si}(a,y)$. Is there a special function for the case $bnot=0$? Any ideas?
calculus integration trigonometry definite-integrals
edited yesterday
rash
20811
asked yesterday
JensJens
458
$endgroup$
1
$begingroup$
What is your question exactly? Even for $b=0$ your integral is not an elementary function of $y$. You will need some special functions here.
$endgroup$
– GReyes
yesterday
$begingroup$
Of course, it will just be the sine integral in the case $b=0$, that is, F(a,0;y) = Si(a,y). My question is if there exists a special function that is the solution of this integral. I will add a line to clarify this.
$endgroup$
– Jens
yesterday
$begingroup$
Ok. Your question is clear now..
$endgroup$
– GReyes
yesterday
add a comment |
$begingroup$
I am interested in this integral (for $a,binmathbb{R}$):
$$ F(a,b;y) := intlimits_0^y text{d}x , frac{sin(ax+b/x)}{x} $$
Usually for integrals of this type it is useful to transform $tilde{x} = 1/x$, but in my case the integral boundaries will get changed as well:
$$ F(a,b;y) := intlimits^infty_{1/y} text{d}tilde{x} , frac{sin(btilde{x}+a/tilde{x})}{tilde{x}} $$
So I conclude that this transformation is only useful for the integral over the entire real line, and not in my case.
In the case of $b=0$ one has of course $F(a,0;y) = text{Si}(a,y)$. Is there a special function for the case $bnot=0$? Any ideas?
calculus integration trigonometry definite-integrals
edited yesterday
rash
20811
asked yesterday
JensJens
458
$endgroup$
I am interested in this integral (for $a,binmathbb{R}$):
$$ F(a,b;y) := intlimits_0^y text{d}x , frac{sin(ax+b/x)}{x} $$
Usually for integrals of this type it is useful to transform $tilde{x} = 1/x$, but in my case the integral boundaries will get changed as well:
$$ F(a,b;y) := intlimits^infty_{1/y} text{d}tilde{x} , frac{sin(btilde{x}+a/tilde{x})}{tilde{x}} $$
So I conclude that this transformation is only useful for the integral over the entire real line, and not in my case.
In the case of $b=0$ one has of course $F(a,0;y) = text{Si}(a,y)$. Is there a special function for the case $bnot=0$? Any ideas?
calculus integration trigonometry definite-integrals
calculus integration trigonometry definite-integrals
edited yesterday
rash
20811
asked yesterday
JensJens
458
edited yesterday
rash
20811
asked yesterday
JensJens
458
edited yesterday
rash
20811
edited yesterday
rash
20811
edited yesterday
rash
20811
20811
asked yesterday
JensJens
458
asked yesterday
JensJens
458
asked yesterday
JensJens
458
458
1
$begingroup$
What is your question exactly? Even for $b=0$ your integral is not an elementary function of $y$. You will need some special functions here.
$endgroup$
– GReyes
yesterday
$begingroup$
Of course, it will just be the sine integral in the case $b=0$, that is, F(a,0;y) = Si(a,y). My question is if there exists a special function that is the solution of this integral. I will add a line to clarify this.
$endgroup$
– Jens
yesterday
$begingroup$
Ok. Your question is clear now..
$endgroup$
– GReyes
yesterday
add a comment |
1
$begingroup$
What is your question exactly? Even for $b=0$ your integral is not an elementary function of $y$. You will need some special functions here.
$endgroup$
– GReyes
yesterday
$begingroup$
Of course, it will just be the sine integral in the case $b=0$, that is, F(a,0;y) = Si(a,y). My question is if there exists a special function that is the solution of this integral. I will add a line to clarify this.
$endgroup$
– Jens
yesterday
$begingroup$
Ok. Your question is clear now..
$endgroup$
– GReyes
yesterday
1
1
$begingroup$
What is your question exactly? Even for $b=0$ your integral is not an elementary function of $y$. You will need some special functions here.
$endgroup$
– GReyes
yesterday
$begingroup$
What is your question exactly? Even for $b=0$ your integral is not an elementary function of $y$. You will need some special functions here.
$endgroup$
– GReyes
yesterday
$begingroup$
Of course, it will just be the sine integral in the case $b=0$, that is, F(a,0;y) = Si(a,y). My question is if there exists a special function that is the solution of this integral. I will add a line to clarify this.
$endgroup$
– Jens
yesterday
$begingroup$
Of course, it will just be the sine integral in the case $b=0$, that is, F(a,0;y) = Si(a,y). My question is if there exists a special function that is the solution of this integral. I will add a line to clarify this.
$endgroup$
– Jens
yesterday
$begingroup$
Ok. Your question is clear now..
$endgroup$
– GReyes
yesterday
$begingroup$
Ok. Your question is clear now..
$endgroup$
– GReyes
yesterday
add a comment |
1 Answer
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$begingroup$
I would enjoy to have a special function for that.
The only thing I have in mind is to simplify the problem using $x=frac y a$ and $c=ab$ to make
$$intfrac{sin left(a x+frac{b}{x}right)}{x},dx=intfrac{sin left(y+frac{c }{y}right)}{y},dy$$ and then expand the integrand as a series built at $c=0$ to get
$$frac{sin left(y+frac{c }{y}right)}{y}=sum_{n=0}^infty frac{ sin left(y+nfrac{pi }{2}right)}{y^{n+1},n!}c^n$$ and then use
$$int frac{sin(y)}{y^k},dy=-frac{1}{2} y^{-k} left((-i y)^k Gamma (1-k,-i y)+(i y)^k Gamma (1-k,i y)right)$$
$$int frac{cos(y)}{y^k},dy=-frac{1}{2} i y^{-k} left((-i y)^k Gamma (1-k,-i y)-(i y)^k Gamma (1-k,i y)right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
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$begingroup$
I would enjoy to have a special function for that.
The only thing I have in mind is to simplify the problem using $x=frac y a$ and $c=ab$ to make
$$intfrac{sin left(a x+frac{b}{x}right)}{x},dx=intfrac{sin left(y+frac{c }{y}right)}{y},dy$$ and then expand the integrand as a series built at $c=0$ to get
$$frac{sin left(y+frac{c }{y}right)}{y}=sum_{n=0}^infty frac{ sin left(y+nfrac{pi }{2}right)}{y^{n+1},n!}c^n$$ and then use
$$int frac{sin(y)}{y^k},dy=-frac{1}{2} y^{-k} left((-i y)^k Gamma (1-k,-i y)+(i y)^k Gamma (1-k,i y)right)$$
$$int frac{cos(y)}{y^k},dy=-frac{1}{2} i y^{-k} left((-i y)^k Gamma (1-k,-i y)-(i y)^k Gamma (1-k,i y)right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
$begingroup$
I would enjoy to have a special function for that.
The only thing I have in mind is to simplify the problem using $x=frac y a$ and $c=ab$ to make
$$intfrac{sin left(a x+frac{b}{x}right)}{x},dx=intfrac{sin left(y+frac{c }{y}right)}{y},dy$$ and then expand the integrand as a series built at $c=0$ to get
$$frac{sin left(y+frac{c }{y}right)}{y}=sum_{n=0}^infty frac{ sin left(y+nfrac{pi }{2}right)}{y^{n+1},n!}c^n$$ and then use
$$int frac{sin(y)}{y^k},dy=-frac{1}{2} y^{-k} left((-i y)^k Gamma (1-k,-i y)+(i y)^k Gamma (1-k,i y)right)$$
$$int frac{cos(y)}{y^k},dy=-frac{1}{2} i y^{-k} left((-i y)^k Gamma (1-k,-i y)-(i y)^k Gamma (1-k,i y)right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
$begingroup$
I would enjoy to have a special function for that.
The only thing I have in mind is to simplify the problem using $x=frac y a$ and $c=ab$ to make
$$intfrac{sin left(a x+frac{b}{x}right)}{x},dx=intfrac{sin left(y+frac{c }{y}right)}{y},dy$$ and then expand the integrand as a series built at $c=0$ to get
$$frac{sin left(y+frac{c }{y}right)}{y}=sum_{n=0}^infty frac{ sin left(y+nfrac{pi }{2}right)}{y^{n+1},n!}c^n$$ and then use
$$int frac{sin(y)}{y^k},dy=-frac{1}{2} y^{-k} left((-i y)^k Gamma (1-k,-i y)+(i y)^k Gamma (1-k,i y)right)$$
$$int frac{cos(y)}{y^k},dy=-frac{1}{2} i y^{-k} left((-i y)^k Gamma (1-k,-i y)-(i y)^k Gamma (1-k,i y)right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
I would enjoy to have a special function for that.
The only thing I have in mind is to simplify the problem using $x=frac y a$ and $c=ab$ to make
$$intfrac{sin left(a x+frac{b}{x}right)}{x},dx=intfrac{sin left(y+frac{c }{y}right)}{y},dy$$ and then expand the integrand as a series built at $c=0$ to get
$$frac{sin left(y+frac{c }{y}right)}{y}=sum_{n=0}^infty frac{ sin left(y+nfrac{pi }{2}right)}{y^{n+1},n!}c^n$$ and then use
$$int frac{sin(y)}{y^k},dy=-frac{1}{2} y^{-k} left((-i y)^k Gamma (1-k,-i y)+(i y)^k Gamma (1-k,i y)right)$$
$$int frac{cos(y)}{y^k},dy=-frac{1}{2} i y^{-k} left((-i y)^k Gamma (1-k,-i y)-(i y)^k Gamma (1-k,i y)right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
123k1157135
answered yesterday
Claude LeiboviciClaude Leibovici
123k1157135
answered yesterday
Claude LeiboviciClaude Leibovici
123k1157135
answered yesterday
Claude LeiboviciClaude Leibovici
123k1157135
123k1157135
add a comment |
add a comment |
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$begingroup$
What is your question exactly? Even for $b=0$ your integral is not an elementary function of $y$. You will need some special functions here.
$endgroup$
– GReyes
yesterday
$begingroup$
Of course, it will just be the sine integral in the case $b=0$, that is, F(a,0;y) = Si(a,y). My question is if there exists a special function that is the solution of this integral. I will add a line to clarify this.
$endgroup$
– Jens
yesterday
$begingroup$
Ok. Your question is clear now..
$endgroup$
– GReyes
yesterday