Finding $int_0^infty frac{arctan(pcdot x)cdot arctan(qcdot x)}{x^2} text{d}t$
$begingroup$
I am attempting to derive the value of the integral
$$
I(p,q)= intlimits_0^infty frac{arctan(pcdot x)cdot arctan(qcdot x)}{x^2} text{d}x
$$
Differentiating the I w.r.t. p and then q gives the expression
$$
frac{partial^2 I}{partial p , partial q} = frac{pi}{2(p+q)}
$$
Now I want to solve this equation but unclear as to how the constant(s) of integration may be found.
pde definite-integrals improper-integrals
edited Dec 6 '18 at 18:22
J.G.
29.1k22845
asked Dec 6 '18 at 17:40
Callie12Callie12
10410
$endgroup$
add a comment |
$begingroup$
I am attempting to derive the value of the integral
$$
I(p,q)= intlimits_0^infty frac{arctan(pcdot x)cdot arctan(qcdot x)}{x^2} text{d}x
$$
Differentiating the I w.r.t. p and then q gives the expression
$$
frac{partial^2 I}{partial p , partial q} = frac{pi}{2(p+q)}
$$
Now I want to solve this equation but unclear as to how the constant(s) of integration may be found.
pde definite-integrals improper-integrals
edited Dec 6 '18 at 18:22
J.G.
29.1k22845
asked Dec 6 '18 at 17:40
Callie12Callie12
10410
$endgroup$
add a comment |
$begingroup$
I am attempting to derive the value of the integral
$$
I(p,q)= intlimits_0^infty frac{arctan(pcdot x)cdot arctan(qcdot x)}{x^2} text{d}x
$$
Differentiating the I w.r.t. p and then q gives the expression
$$
frac{partial^2 I}{partial p , partial q} = frac{pi}{2(p+q)}
$$
Now I want to solve this equation but unclear as to how the constant(s) of integration may be found.
pde definite-integrals improper-integrals
edited Dec 6 '18 at 18:22
J.G.
29.1k22845
asked Dec 6 '18 at 17:40
Callie12Callie12
10410
$endgroup$
I am attempting to derive the value of the integral
$$
I(p,q)= intlimits_0^infty frac{arctan(pcdot x)cdot arctan(qcdot x)}{x^2} text{d}x
$$
Differentiating the I w.r.t. p and then q gives the expression
$$
frac{partial^2 I}{partial p , partial q} = frac{pi}{2(p+q)}
$$
Now I want to solve this equation but unclear as to how the constant(s) of integration may be found.
pde definite-integrals improper-integrals
pde definite-integrals improper-integrals
edited Dec 6 '18 at 18:22
J.G.
29.1k22845
asked Dec 6 '18 at 17:40
Callie12Callie12
10410
edited Dec 6 '18 at 18:22
J.G.
29.1k22845
asked Dec 6 '18 at 17:40
Callie12Callie12
10410
edited Dec 6 '18 at 18:22
J.G.
29.1k22845
edited Dec 6 '18 at 18:22
J.G.
29.1k22845
edited Dec 6 '18 at 18:22
J.G.
29.1k22845
29.1k22845
asked Dec 6 '18 at 17:40
Callie12Callie12
10410
asked Dec 6 '18 at 17:40
Callie12Callie12
10410
asked Dec 6 '18 at 17:40
Callie12Callie12
10410
10410
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Use the fact that $I(p,0) = I(0,q) = 0$.
answered Dec 6 '18 at 17:55
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
Plus the derivatives vanish when an argument is $0$.
$endgroup$
– J.G.
Dec 6 '18 at 18:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use the fact that $I(p,0) = I(0,q) = 0$.
answered Dec 6 '18 at 17:55
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
Plus the derivatives vanish when an argument is $0$.
$endgroup$
– J.G.
Dec 6 '18 at 18:23
add a comment |
$begingroup$
Use the fact that $I(p,0) = I(0,q) = 0$.
answered Dec 6 '18 at 17:55
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
Plus the derivatives vanish when an argument is $0$.
$endgroup$
– J.G.
Dec 6 '18 at 18:23
add a comment |
$begingroup$
Use the fact that $I(p,0) = I(0,q) = 0$.
answered Dec 6 '18 at 17:55
Robert IsraelRobert Israel
326k23215469
$endgroup$
Use the fact that $I(p,0) = I(0,q) = 0$.
answered Dec 6 '18 at 17:55
Robert IsraelRobert Israel
326k23215469
answered Dec 6 '18 at 17:55
Robert IsraelRobert Israel
326k23215469
answered Dec 6 '18 at 17:55
Robert IsraelRobert Israel
326k23215469
answered Dec 6 '18 at 17:55
Robert IsraelRobert Israel
326k23215469
326k23215469
$begingroup$
Plus the derivatives vanish when an argument is $0$.
$endgroup$
– J.G.
Dec 6 '18 at 18:23
add a comment |
$begingroup$
Plus the derivatives vanish when an argument is $0$.
$endgroup$
– J.G.
Dec 6 '18 at 18:23
$begingroup$
Plus the derivatives vanish when an argument is $0$.
$endgroup$
– J.G.
Dec 6 '18 at 18:23
$begingroup$
Plus the derivatives vanish when an argument is $0$.
$endgroup$
– J.G.
Dec 6 '18 at 18:23
add a comment |
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