Compute log r dr 1+x²
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter0: A Review Of Basic Algebra
Section0.CT: Chapter Test
Problem 34CT
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[Complex Variables] How do you solve this? Thanks
Hint: Use the second photo as reference since the question is similar
Transcribed Image Text:Compute
log r
dx
1+x²
3.
![Example11.4 Functions with Branch Cut Compute
log a
Solution
After we choose this branch cut, set the principle branch of log z = In|z| + iarg(2)
with arg(2) E (-3, ). Consider the function
log z
f(2) =
(1+ 2²)²
Im
Y2
Y4
-R
R Re
Y3
Y1
Figure 11.5: Half-Donut Contour
Now z = i is the only pole in the interior of the contour, with order equal to 2. It follows that
H(2)
log z
f(2) =
H(2)
(z – i)²'
(z + i)²
Hence
1
log z
Ti 1
+2-
4i
1
i
Res(f, i) = H'(i)
|3D
z(z+i)²
2 8i
4
8.
z=i
$(4 + 2),
Now
log t
(1 +t2)2 dt;
log (Ret)
(1+ R²c2it)2
f(2)dz
f(2)dz = /
iRe" dt;
logt + in
dtB
f(2)dz = ,(1+t2)²
log (cei(x-t))
f(2)dz = T+c²c2i(x-t))2™
It follows that, for R>0 large enough,
log (Ret)
siRe" dt < R|
log (Ret)
f(z)dz
dt
(1+ R?c2it)2
2nRln R R→0
→ 0;
(R2 – 1)2
log (ce'(*-t))
(1+e²c2i(a–t))2“
log (cʻ(r-t)
-t),
f(z)dz
-i)dt
dt
(1+e²c2i(x-t))²
2 In e
dt
1– 2
<€
0;
Im
Y2
-R
R
Re
Y3
Y1
where in the last line of the above inequality we have used the fact that for small e> 0 and large
R>0,
| In R+ log e“| = | In R+ In |e“| + iarg(c“)|
|In R+0+ it| < |In R| + |t| < 2 ln R;
| log Re"|
%3D
|1 + R°c2] >| \R?c2"| – 1|2 R² – 1
| log(ce(r-)| = | In c + log e'("-1)| = | Inc + (In |e(#-4)| + iarg(e'(a-t)|
„i(n-
I Ine +0+i(π - ) < |Ine|+|ㅠ -1<| Inel + π < 2| Ine
|1+?e2i(x-t)| >| 1- |2e2i(x-1)||=1-2
Now we should estimate the other two curve integrals.
log t
f(z)dz =
we keep this quantity as it is. Next
| {(2)dz = [ * log |le| + in
(1+t²)²
23
It turns out that
| {(:)dz +
f(2)dz
73
log |t|
dt +
(1+t²)²
log t
1
%3D
(1+t2)2 dt + in
R (1+t²)2dt
R
R
log t
1
=2
dt + in
(1+t²)2"
(1+t²)²
As a result, we have actually obtained that
= 2niRes(f,i)
4
= | f(2)dz + f(2)dz + | f(2)dz+
f(2)dz
Y3
R
log t
dt + in
(1+t²)²
1
dt+
| f(2)dz+
= 2
f(2)dz
(1+t²)2
On the right hand side, pass R→ 0 ande → 0, we see that
log t
1
dt
(1+t²)²
2
2
(1+t2)2 dt + in
Compare the real part and the imaginary part respectively, we see that
log t
(1 + t²)2 dt:
Remark Indeed, as a by-product, we also get that (by comparing the imaginary parts)
1
(1+t²)2 at](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddb3e5af-ed88-460f-addc-a49c08dc1c74%2Fd7299821-6576-44b0-9236-44038e9e2636%2Fivos808_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example11.4 Functions with Branch Cut Compute
log a
Solution
After we choose this branch cut, set the principle branch of log z = In|z| + iarg(2)
with arg(2) E (-3, ). Consider the function
log z
f(2) =
(1+ 2²)²
Im
Y2
Y4
-R
R Re
Y3
Y1
Figure 11.5: Half-Donut Contour
Now z = i is the only pole in the interior of the contour, with order equal to 2. It follows that
H(2)
log z
f(2) =
H(2)
(z – i)²'
(z + i)²
Hence
1
log z
Ti 1
+2-
4i
1
i
Res(f, i) = H'(i)
|3D
z(z+i)²
2 8i
4
8.
z=i
$(4 + 2),
Now
log t
(1 +t2)2 dt;
log (Ret)
(1+ R²c2it)2
f(2)dz
f(2)dz = /
iRe" dt;
logt + in
dtB
f(2)dz = ,(1+t2)²
log (cei(x-t))
f(2)dz = T+c²c2i(x-t))2™
It follows that, for R>0 large enough,
log (Ret)
siRe" dt < R|
log (Ret)
f(z)dz
dt
(1+ R?c2it)2
2nRln R R→0
→ 0;
(R2 – 1)2
log (ce'(*-t))
(1+e²c2i(a–t))2“
log (cʻ(r-t)
-t),
f(z)dz
-i)dt
dt
(1+e²c2i(x-t))²
2 In e
dt
1– 2
<€
0;
Im
Y2
-R
R
Re
Y3
Y1
where in the last line of the above inequality we have used the fact that for small e> 0 and large
R>0,
| In R+ log e“| = | In R+ In |e“| + iarg(c“)|
|In R+0+ it| < |In R| + |t| < 2 ln R;
| log Re"|
%3D
|1 + R°c2] >| \R?c2"| – 1|2 R² – 1
| log(ce(r-)| = | In c + log e'("-1)| = | Inc + (In |e(#-4)| + iarg(e'(a-t)|
„i(n-
I Ine +0+i(π - ) < |Ine|+|ㅠ -1<| Inel + π < 2| Ine
|1+?e2i(x-t)| >| 1- |2e2i(x-1)||=1-2
Now we should estimate the other two curve integrals.
log t
f(z)dz =
we keep this quantity as it is. Next
| {(2)dz = [ * log |le| + in
(1+t²)²
23
It turns out that
| {(:)dz +
f(2)dz
73
log |t|
dt +
(1+t²)²
log t
1
%3D
(1+t2)2 dt + in
R (1+t²)2dt
R
R
log t
1
=2
dt + in
(1+t²)2"
(1+t²)²
As a result, we have actually obtained that
= 2niRes(f,i)
4
= | f(2)dz + f(2)dz + | f(2)dz+
f(2)dz
Y3
R
log t
dt + in
(1+t²)²
1
dt+
| f(2)dz+
= 2
f(2)dz
(1+t²)2
On the right hand side, pass R→ 0 ande → 0, we see that
log t
1
dt
(1+t²)²
2
2
(1+t2)2 dt + in
Compare the real part and the imaginary part respectively, we see that
log t
(1 + t²)2 dt:
Remark Indeed, as a by-product, we also get that (by comparing the imaginary parts)
1
(1+t²)2 at
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