Why does $int_1^sqrt2 frac{1}{x}lnleft(frac{2-2x^2+x^4}{2x-2x^2+x^3}right)dx$ equal to $0$?












23












$begingroup$


In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.



This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.



Any ideas?



EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$






integration definite-integrals logarithms symmetry







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$endgroup$








  • 1




    $begingroup$
    @Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
    $endgroup$
    – Frpzzd
    Nov 20 '18 at 0:47
















23












$begingroup$


In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.



This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.



Any ideas?



EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$






integration definite-integrals logarithms symmetry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
    $endgroup$
    – Frpzzd
    Nov 20 '18 at 0:47














23












23








23


8



$begingroup$


In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.



This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.



Any ideas?



EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$






integration definite-integrals logarithms symmetry







share|cite|improve this question











$endgroup$




In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.



This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.



Any ideas?



EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$






integration definite-integrals logarithms symmetry




integration definite-integrals logarithms symmetry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 12:44









Zacky

7,3601961




7,3601961










asked Nov 20 '18 at 0:42









FrpzzdFrpzzd

23k841109




23k841109








  • 1




    $begingroup$
    @Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
    $endgroup$
    – Frpzzd
    Nov 20 '18 at 0:47














  • 1




    $begingroup$
    @Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
    $endgroup$
    – Frpzzd
    Nov 20 '18 at 0:47








1




1




$begingroup$
@Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
$endgroup$
– Frpzzd
Nov 20 '18 at 0:47




$begingroup$
@Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
$endgroup$
– Frpzzd
Nov 20 '18 at 0:47










1 Answer
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13












$begingroup$

We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=int_1^sqrt2 frac{1}{x}lnbigg(frac{x^4-2x^2+2}{x^2-2x+2}bigg)dx=color{green}{int_1^sqrt 2frac{1}{x}ln x dx=frac18ln^2 2}$$
First let us take the LHS integral and split it in two parts:
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt2 frac{ln(x^2-2x+2)}{x}dx}$$
For the second one $(I_2)$ we will substitute $displaystyle{x=frac{2}{t}rightarrow dx=-frac{2}{t^2}dt}$:
$$I_2=int_{sqrt 2}^2 frac{lnleft(frac{2(t^2-2t+2)}{t^2}right)}{frac{2}{t}}frac{2}{t^2}dtoverset{t=x}=int_{sqrt 2}^2 frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx$$
Adding with the original $I_2$ leads to:
$${2I_2=int_1^2frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx}$$
$${Rightarrow I_2=frac12int_1^2 frac{ln(x^2-2x+2)}{x}dx-frac{1}{8}ln^22}overset{x=t^2}=color{blue}{int_1^sqrt{2}frac{ln(t^4-2t^2+2)}{t}dt-frac{1}{8}ln^2 2}$$
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt{2}frac{ln(x^4-2x^2+2)}{x}dx+frac{1}{8}ln^2 2}=color{green}{frac18ln^2 2}$$




Your conjecture is indeed also correct since by the exact same method we can show that: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{t-sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx=frac{1}{8}ln^2t$$
And this time after we split the LHS integral into two parts, we will substitute in the second integral $displaystyle{x=frac{t}{y}}$ ($t$ is a constant here),
followed by an addition with the original $I_2$ from there and the result follows. Of course it is valid in the following form too: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t+sx^2+x^4}{t+sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Amazing, you did it with nothing more than simple substitutions! Could you take a look at this question as well, if it interests you: math.stackexchange.com/questions/3007841/…
    $endgroup$
    – Frpzzd
    Dec 22 '18 at 14:38






  • 1




    $begingroup$
    OMG, this is very clever way! +1
    $endgroup$
    – Anastasiya-Romanova 秀
    Dec 22 '18 at 15:33











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13












$begingroup$

We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=int_1^sqrt2 frac{1}{x}lnbigg(frac{x^4-2x^2+2}{x^2-2x+2}bigg)dx=color{green}{int_1^sqrt 2frac{1}{x}ln x dx=frac18ln^2 2}$$
First let us take the LHS integral and split it in two parts:
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt2 frac{ln(x^2-2x+2)}{x}dx}$$
For the second one $(I_2)$ we will substitute $displaystyle{x=frac{2}{t}rightarrow dx=-frac{2}{t^2}dt}$:
$$I_2=int_{sqrt 2}^2 frac{lnleft(frac{2(t^2-2t+2)}{t^2}right)}{frac{2}{t}}frac{2}{t^2}dtoverset{t=x}=int_{sqrt 2}^2 frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx$$
Adding with the original $I_2$ leads to:
$${2I_2=int_1^2frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx}$$
$${Rightarrow I_2=frac12int_1^2 frac{ln(x^2-2x+2)}{x}dx-frac{1}{8}ln^22}overset{x=t^2}=color{blue}{int_1^sqrt{2}frac{ln(t^4-2t^2+2)}{t}dt-frac{1}{8}ln^2 2}$$
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt{2}frac{ln(x^4-2x^2+2)}{x}dx+frac{1}{8}ln^2 2}=color{green}{frac18ln^2 2}$$




Your conjecture is indeed also correct since by the exact same method we can show that: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{t-sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx=frac{1}{8}ln^2t$$
And this time after we split the LHS integral into two parts, we will substitute in the second integral $displaystyle{x=frac{t}{y}}$ ($t$ is a constant here),
followed by an addition with the original $I_2$ from there and the result follows. Of course it is valid in the following form too: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t+sx^2+x^4}{t+sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Amazing, you did it with nothing more than simple substitutions! Could you take a look at this question as well, if it interests you: math.stackexchange.com/questions/3007841/…
    $endgroup$
    – Frpzzd
    Dec 22 '18 at 14:38






  • 1




    $begingroup$
    OMG, this is very clever way! +1
    $endgroup$
    – Anastasiya-Romanova 秀
    Dec 22 '18 at 15:33
















13












$begingroup$

We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=int_1^sqrt2 frac{1}{x}lnbigg(frac{x^4-2x^2+2}{x^2-2x+2}bigg)dx=color{green}{int_1^sqrt 2frac{1}{x}ln x dx=frac18ln^2 2}$$
First let us take the LHS integral and split it in two parts:
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt2 frac{ln(x^2-2x+2)}{x}dx}$$
For the second one $(I_2)$ we will substitute $displaystyle{x=frac{2}{t}rightarrow dx=-frac{2}{t^2}dt}$:
$$I_2=int_{sqrt 2}^2 frac{lnleft(frac{2(t^2-2t+2)}{t^2}right)}{frac{2}{t}}frac{2}{t^2}dtoverset{t=x}=int_{sqrt 2}^2 frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx$$
Adding with the original $I_2$ leads to:
$${2I_2=int_1^2frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx}$$
$${Rightarrow I_2=frac12int_1^2 frac{ln(x^2-2x+2)}{x}dx-frac{1}{8}ln^22}overset{x=t^2}=color{blue}{int_1^sqrt{2}frac{ln(t^4-2t^2+2)}{t}dt-frac{1}{8}ln^2 2}$$
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt{2}frac{ln(x^4-2x^2+2)}{x}dx+frac{1}{8}ln^2 2}=color{green}{frac18ln^2 2}$$




Your conjecture is indeed also correct since by the exact same method we can show that: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{t-sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx=frac{1}{8}ln^2t$$
And this time after we split the LHS integral into two parts, we will substitute in the second integral $displaystyle{x=frac{t}{y}}$ ($t$ is a constant here),
followed by an addition with the original $I_2$ from there and the result follows. Of course it is valid in the following form too: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t+sx^2+x^4}{t+sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Amazing, you did it with nothing more than simple substitutions! Could you take a look at this question as well, if it interests you: math.stackexchange.com/questions/3007841/…
    $endgroup$
    – Frpzzd
    Dec 22 '18 at 14:38






  • 1




    $begingroup$
    OMG, this is very clever way! +1
    $endgroup$
    – Anastasiya-Romanova 秀
    Dec 22 '18 at 15:33














13












13








13





$begingroup$

We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=int_1^sqrt2 frac{1}{x}lnbigg(frac{x^4-2x^2+2}{x^2-2x+2}bigg)dx=color{green}{int_1^sqrt 2frac{1}{x}ln x dx=frac18ln^2 2}$$
First let us take the LHS integral and split it in two parts:
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt2 frac{ln(x^2-2x+2)}{x}dx}$$
For the second one $(I_2)$ we will substitute $displaystyle{x=frac{2}{t}rightarrow dx=-frac{2}{t^2}dt}$:
$$I_2=int_{sqrt 2}^2 frac{lnleft(frac{2(t^2-2t+2)}{t^2}right)}{frac{2}{t}}frac{2}{t^2}dtoverset{t=x}=int_{sqrt 2}^2 frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx$$
Adding with the original $I_2$ leads to:
$${2I_2=int_1^2frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx}$$
$${Rightarrow I_2=frac12int_1^2 frac{ln(x^2-2x+2)}{x}dx-frac{1}{8}ln^22}overset{x=t^2}=color{blue}{int_1^sqrt{2}frac{ln(t^4-2t^2+2)}{t}dt-frac{1}{8}ln^2 2}$$
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt{2}frac{ln(x^4-2x^2+2)}{x}dx+frac{1}{8}ln^2 2}=color{green}{frac18ln^2 2}$$




Your conjecture is indeed also correct since by the exact same method we can show that: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{t-sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx=frac{1}{8}ln^2t$$
And this time after we split the LHS integral into two parts, we will substitute in the second integral $displaystyle{x=frac{t}{y}}$ ($t$ is a constant here),
followed by an addition with the original $I_2$ from there and the result follows. Of course it is valid in the following form too: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t+sx^2+x^4}{t+sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx$$






share|cite|improve this answer











$endgroup$



We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=int_1^sqrt2 frac{1}{x}lnbigg(frac{x^4-2x^2+2}{x^2-2x+2}bigg)dx=color{green}{int_1^sqrt 2frac{1}{x}ln x dx=frac18ln^2 2}$$
First let us take the LHS integral and split it in two parts:
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt2 frac{ln(x^2-2x+2)}{x}dx}$$
For the second one $(I_2)$ we will substitute $displaystyle{x=frac{2}{t}rightarrow dx=-frac{2}{t^2}dt}$:
$$I_2=int_{sqrt 2}^2 frac{lnleft(frac{2(t^2-2t+2)}{t^2}right)}{frac{2}{t}}frac{2}{t^2}dtoverset{t=x}=int_{sqrt 2}^2 frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx$$
Adding with the original $I_2$ leads to:
$${2I_2=int_1^2frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx}$$
$${Rightarrow I_2=frac12int_1^2 frac{ln(x^2-2x+2)}{x}dx-frac{1}{8}ln^22}overset{x=t^2}=color{blue}{int_1^sqrt{2}frac{ln(t^4-2t^2+2)}{t}dt-frac{1}{8}ln^2 2}$$
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt{2}frac{ln(x^4-2x^2+2)}{x}dx+frac{1}{8}ln^2 2}=color{green}{frac18ln^2 2}$$




Your conjecture is indeed also correct since by the exact same method we can show that: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{t-sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx=frac{1}{8}ln^2t$$
And this time after we split the LHS integral into two parts, we will substitute in the second integral $displaystyle{x=frac{t}{y}}$ ($t$ is a constant here),
followed by an addition with the original $I_2$ from there and the result follows. Of course it is valid in the following form too: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t+sx^2+x^4}{t+sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 19:43

























answered Dec 22 '18 at 0:38









ZackyZacky

7,3601961




7,3601961








  • 1




    $begingroup$
    Amazing, you did it with nothing more than simple substitutions! Could you take a look at this question as well, if it interests you: math.stackexchange.com/questions/3007841/…
    $endgroup$
    – Frpzzd
    Dec 22 '18 at 14:38






  • 1




    $begingroup$
    OMG, this is very clever way! +1
    $endgroup$
    – Anastasiya-Romanova 秀
    Dec 22 '18 at 15:33














  • 1




    $begingroup$
    Amazing, you did it with nothing more than simple substitutions! Could you take a look at this question as well, if it interests you: math.stackexchange.com/questions/3007841/…
    $endgroup$
    – Frpzzd
    Dec 22 '18 at 14:38






  • 1




    $begingroup$
    OMG, this is very clever way! +1
    $endgroup$
    – Anastasiya-Romanova 秀
    Dec 22 '18 at 15:33








1




1




$begingroup$
Amazing, you did it with nothing more than simple substitutions! Could you take a look at this question as well, if it interests you: math.stackexchange.com/questions/3007841/…
$endgroup$
– Frpzzd
Dec 22 '18 at 14:38




$begingroup$
Amazing, you did it with nothing more than simple substitutions! Could you take a look at this question as well, if it interests you: math.stackexchange.com/questions/3007841/…
$endgroup$
– Frpzzd
Dec 22 '18 at 14:38




1




1




$begingroup$
OMG, this is very clever way! +1
$endgroup$
– Anastasiya-Romanova 秀
Dec 22 '18 at 15:33




$begingroup$
OMG, this is very clever way! +1
$endgroup$
– Anastasiya-Romanova 秀
Dec 22 '18 at 15:33


















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Wiesbaden

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Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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