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Is there a rational function $f(x,a)$ such that $int_0^1 f(x,a) dx = fracln(a)a$
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I was wondering if a rational function (quotient between polynomials) $f(x,a)$ exists such that $int_0^1 f(x,a) dx = fracln(a)a.$
For example, I could find, while playing with wolfram, a not rational $f(x,a) =frac ln(ax)+1a$ that integrated on $x$ from $0$ to $1$ yelds $fracln aa$. But the function not being rational does not really help me. How to proceed on a problem like this?
calculus integration definite-integrals
asked Jul 15 at 1:27
Pinteco
446110
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up vote
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I was wondering if a rational function (quotient between polynomials) $f(x,a)$ exists such that $int_0^1 f(x,a) dx = fracln(a)a.$
For example, I could find, while playing with wolfram, a not rational $f(x,a) =frac ln(ax)+1a$ that integrated on $x$ from $0$ to $1$ yelds $fracln aa$. But the function not being rational does not really help me. How to proceed on a problem like this?
calculus integration definite-integrals
asked Jul 15 at 1:27
Pinteco
446110
This is equivalent to finding an $F(x, a)$ such that $F(1,a) - F(0,a) = ln(a)/a$. Where F' is rational.
– Kaynex
Jul 15 at 1:46
@Kaynex What does "$F'$ rational" mean in this multivariable case? $fracpartial Fpartial x$?
– Randall
Jul 15 at 2:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was wondering if a rational function (quotient between polynomials) $f(x,a)$ exists such that $int_0^1 f(x,a) dx = fracln(a)a.$
For example, I could find, while playing with wolfram, a not rational $f(x,a) =frac ln(ax)+1a$ that integrated on $x$ from $0$ to $1$ yelds $fracln aa$. But the function not being rational does not really help me. How to proceed on a problem like this?
calculus integration definite-integrals
asked Jul 15 at 1:27
Pinteco
446110
I was wondering if a rational function (quotient between polynomials) $f(x,a)$ exists such that $int_0^1 f(x,a) dx = fracln(a)a.$
For example, I could find, while playing with wolfram, a not rational $f(x,a) =frac ln(ax)+1a$ that integrated on $x$ from $0$ to $1$ yelds $fracln aa$. But the function not being rational does not really help me. How to proceed on a problem like this?
calculus integration definite-integrals
asked Jul 15 at 1:27
Pinteco
446110
asked Jul 15 at 1:27
Pinteco
446110
asked Jul 15 at 1:27
Pinteco
446110
asked Jul 15 at 1:27
Pinteco
446110
446110
This is equivalent to finding an $F(x, a)$ such that $F(1,a) - F(0,a) = ln(a)/a$. Where F' is rational.
– Kaynex
Jul 15 at 1:46
@Kaynex What does "$F'$ rational" mean in this multivariable case? $fracpartial Fpartial x$?
– Randall
Jul 15 at 2:08
add a comment |Â
This is equivalent to finding an $F(x, a)$ such that $F(1,a) - F(0,a) = ln(a)/a$. Where F' is rational.
– Kaynex
Jul 15 at 1:46
@Kaynex What does "$F'$ rational" mean in this multivariable case? $fracpartial Fpartial x$?
– Randall
Jul 15 at 2:08
This is equivalent to finding an $F(x, a)$ such that $F(1,a) - F(0,a) = ln(a)/a$. Where F' is rational.
– Kaynex
Jul 15 at 1:46
This is equivalent to finding an $F(x, a)$ such that $F(1,a) - F(0,a) = ln(a)/a$. Where F' is rational.
– Kaynex
Jul 15 at 1:46
@Kaynex What does "$F'$ rational" mean in this multivariable case? $fracpartial Fpartial x$?
– Randall
Jul 15 at 2:08
@Kaynex What does "$F'$ rational" mean in this multivariable case? $fracpartial Fpartial x$?
– Randall
Jul 15 at 2:08
add a comment |Â
1 Answer
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Doesn't $f(x,a) = fraca-1a frac1(a-1)x+1$ work? Assume $a >0$ throughout (otherwise your log term makes no sense).
Let $g(x)= (a-1)x+1$. This is clearly monotone increasing or decreasing as long as $a neq 1$. As $g(0)=1$ and $g(1) = a$, $g(x)$ is never $0$ on $[0,1]$ by the IVT. If $a=1$ then $g(x) equiv 1$ and again $g$ is never $0$. So, either way, $f(x,a) = fraca-1a frac1(a-1)x+1$ has no discontinuities w/r/t $x$ over $[0,1]$.
You can now check---unless I did something really wrong---that $$fraca-1a int_0^1 frac1(a-1)x+1 dx = fracln aa.$$
To answer your question how to proceed on a problem like this?, I don't know. I just experimented and made adjustments as I went, starting with $int_1^a frac1x dx = ln (a)$, then dividing by $a$, then scooting the interval down to $[0,1]$, etc. Nothing clever.
answered Jul 15 at 2:17
Randall
7,2471825
1
Look right. Wolfy agrees. Good work.
– marty cohen
Jul 15 at 3:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Doesn't $f(x,a) = fraca-1a frac1(a-1)x+1$ work? Assume $a >0$ throughout (otherwise your log term makes no sense).
Let $g(x)= (a-1)x+1$. This is clearly monotone increasing or decreasing as long as $a neq 1$. As $g(0)=1$ and $g(1) = a$, $g(x)$ is never $0$ on $[0,1]$ by the IVT. If $a=1$ then $g(x) equiv 1$ and again $g$ is never $0$. So, either way, $f(x,a) = fraca-1a frac1(a-1)x+1$ has no discontinuities w/r/t $x$ over $[0,1]$.
You can now check---unless I did something really wrong---that $$fraca-1a int_0^1 frac1(a-1)x+1 dx = fracln aa.$$
To answer your question how to proceed on a problem like this?, I don't know. I just experimented and made adjustments as I went, starting with $int_1^a frac1x dx = ln (a)$, then dividing by $a$, then scooting the interval down to $[0,1]$, etc. Nothing clever.
answered Jul 15 at 2:17
Randall
7,2471825
1
Look right. Wolfy agrees. Good work.
– marty cohen
Jul 15 at 3:51
add a comment |Â
up vote
2
down vote
accepted
Doesn't $f(x,a) = fraca-1a frac1(a-1)x+1$ work? Assume $a >0$ throughout (otherwise your log term makes no sense).
Let $g(x)= (a-1)x+1$. This is clearly monotone increasing or decreasing as long as $a neq 1$. As $g(0)=1$ and $g(1) = a$, $g(x)$ is never $0$ on $[0,1]$ by the IVT. If $a=1$ then $g(x) equiv 1$ and again $g$ is never $0$. So, either way, $f(x,a) = fraca-1a frac1(a-1)x+1$ has no discontinuities w/r/t $x$ over $[0,1]$.
You can now check---unless I did something really wrong---that $$fraca-1a int_0^1 frac1(a-1)x+1 dx = fracln aa.$$
To answer your question how to proceed on a problem like this?, I don't know. I just experimented and made adjustments as I went, starting with $int_1^a frac1x dx = ln (a)$, then dividing by $a$, then scooting the interval down to $[0,1]$, etc. Nothing clever.
answered Jul 15 at 2:17
Randall
7,2471825
1
Look right. Wolfy agrees. Good work.
– marty cohen
Jul 15 at 3:51
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Doesn't $f(x,a) = fraca-1a frac1(a-1)x+1$ work? Assume $a >0$ throughout (otherwise your log term makes no sense).
Let $g(x)= (a-1)x+1$. This is clearly monotone increasing or decreasing as long as $a neq 1$. As $g(0)=1$ and $g(1) = a$, $g(x)$ is never $0$ on $[0,1]$ by the IVT. If $a=1$ then $g(x) equiv 1$ and again $g$ is never $0$. So, either way, $f(x,a) = fraca-1a frac1(a-1)x+1$ has no discontinuities w/r/t $x$ over $[0,1]$.
You can now check---unless I did something really wrong---that $$fraca-1a int_0^1 frac1(a-1)x+1 dx = fracln aa.$$
To answer your question how to proceed on a problem like this?, I don't know. I just experimented and made adjustments as I went, starting with $int_1^a frac1x dx = ln (a)$, then dividing by $a$, then scooting the interval down to $[0,1]$, etc. Nothing clever.
answered Jul 15 at 2:17
Randall
7,2471825
Doesn't $f(x,a) = fraca-1a frac1(a-1)x+1$ work? Assume $a >0$ throughout (otherwise your log term makes no sense).
Let $g(x)= (a-1)x+1$. This is clearly monotone increasing or decreasing as long as $a neq 1$. As $g(0)=1$ and $g(1) = a$, $g(x)$ is never $0$ on $[0,1]$ by the IVT. If $a=1$ then $g(x) equiv 1$ and again $g$ is never $0$. So, either way, $f(x,a) = fraca-1a frac1(a-1)x+1$ has no discontinuities w/r/t $x$ over $[0,1]$.
You can now check---unless I did something really wrong---that $$fraca-1a int_0^1 frac1(a-1)x+1 dx = fracln aa.$$
To answer your question how to proceed on a problem like this?, I don't know. I just experimented and made adjustments as I went, starting with $int_1^a frac1x dx = ln (a)$, then dividing by $a$, then scooting the interval down to $[0,1]$, etc. Nothing clever.
answered Jul 15 at 2:17
Randall
7,2471825
edited Jul 15 at 2:27
answered Jul 15 at 2:17
Randall
7,2471825
answered Jul 15 at 2:17
Randall
7,2471825
answered Jul 15 at 2:17
Randall
7,2471825
7,2471825
1
Look right. Wolfy agrees. Good work.
– marty cohen
Jul 15 at 3:51
add a comment |Â
1
Look right. Wolfy agrees. Good work.
– marty cohen
Jul 15 at 3:51
1
1
Look right. Wolfy agrees. Good work.
– marty cohen
Jul 15 at 3:51
Look right. Wolfy agrees. Good work.
– marty cohen
Jul 15 at 3:51
add a comment |Â
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This is equivalent to finding an $F(x, a)$ such that $F(1,a) - F(0,a) = ln(a)/a$. Where F' is rational.
– Kaynex
Jul 15 at 1:46
@Kaynex What does "$F'$ rational" mean in this multivariable case? $fracpartial Fpartial x$?
– Randall
Jul 15 at 2:08