Find using Residue Theorem the following integral












0












$begingroup$


Find using Residue Theorem



$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



My try:



I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.



and $C_2$ is the line joining $-1$ to $1$.



I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.



Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Any help from someone here?



Thanks for reading my post






complex-analysis residue-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 17:02












  • $begingroup$
    @DonAntonio;edited my question,can u have a look now
    $endgroup$
    – Join_PhD
    Dec 4 '18 at 17:07










  • $begingroup$
    Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 18:14


















0












$begingroup$


Find using Residue Theorem



$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



My try:



I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.



and $C_2$ is the line joining $-1$ to $1$.



I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.



Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Any help from someone here?



Thanks for reading my post






complex-analysis residue-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 17:02












  • $begingroup$
    @DonAntonio;edited my question,can u have a look now
    $endgroup$
    – Join_PhD
    Dec 4 '18 at 17:07










  • $begingroup$
    Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 18:14
















0












0








0


0



$begingroup$


Find using Residue Theorem



$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



My try:



I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.



and $C_2$ is the line joining $-1$ to $1$.



I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.



Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Any help from someone here?



Thanks for reading my post






complex-analysis residue-calculus







share|cite|improve this question











$endgroup$




Find using Residue Theorem



$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



My try:



I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.



and $C_2$ is the line joining $-1$ to $1$.



I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.



Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Any help from someone here?



Thanks for reading my post






complex-analysis residue-calculus




complex-analysis residue-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 17:06







Join_PhD

















asked Dec 4 '18 at 16:57









Join_PhDJoin_PhD

3378




3378












  • $begingroup$
    Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 17:02












  • $begingroup$
    @DonAntonio;edited my question,can u have a look now
    $endgroup$
    – Join_PhD
    Dec 4 '18 at 17:07










  • $begingroup$
    Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 18:14




















  • $begingroup$
    Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 17:02












  • $begingroup$
    @DonAntonio;edited my question,can u have a look now
    $endgroup$
    – Join_PhD
    Dec 4 '18 at 17:07










  • $begingroup$
    Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 18:14


















$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02






$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02














$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07




$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07












$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14






$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14












1 Answer
1






active

oldest

votes


















0












$begingroup$

The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.






share|cite|improve this answer









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  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09
















0












$begingroup$

The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09














0












0








0





$begingroup$

The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.






share|cite|improve this answer









$endgroup$



The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.







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share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 10:07









DylanDylan

12.4k31026




12.4k31026












  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09


















  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09
















$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09




$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09


















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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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