Sorry for my English if there is any mistake. The exercice for which I need help is the following:

Compute using complex methods: $I=\int_1 ^\infty \frac{\mathrm{d}x}{x^2+1}$

i) Choose the complex function to integrate.

I guess it is $f(z)=1/(z^2+1)$

ii) Choose the contour.

I don't know what to do here. In my notes there are only examples when the integral is from $-\infty$ to $\infty$, so it takes a circumference of radius $r$ and lets it tend to $\infty$.

iii) Compute the integrals along circumferences.

iv) Compute the branch cut.

I don't know why is this question here, because the function is not multivalued.

v) Compute the integral.

vi) Compute the integral using elemental methods.

$I=\int_1 ^\infty \frac{\mathrm{d}x}{x^2+1}=\lim _{a\rightarrow \infty} \int _1 ^a \frac{\mathrm{d}x}{x^2+1}= \lim _{a\rightarrow \infty} \left[ \arctan x \right]_1 ^a =\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$

Edit: The answer might follow the steps given. My teacher did an exercice that way, but I don't know why he uses such method (the example is in a comment within the answers).


Note that with $x = 1/u$ $$\int_0^1 \dfrac{dx}{x^2+1} = \int_\infty^1 \dfrac{-du/u^2}{1/u^2+1} = \int_1^\infty \dfrac{du}{1+u^2}$$ so your integral is $\displaystyle \dfrac{1}{2}\int_0^\infty \dfrac{dx}{x^2+1} = \dfrac{1}{4} \int_{-\infty}^\infty \dfrac{dx}{x^2+1} $. Now use the semicircular contour as in DonAntonio's answer.


What about taking the semicircle $$\Gamma:=[-R,R]\cup\gamma_R:=\{z\;\;:\;\;|z|= R\,\,,\,0<\arg z< \pi\}\,\,,\,R>1$$

Note that within this closed path there's only one singularity of the function, the simple pole $\,z=i\,$, and the residue of $\,\displaystyle{f(z)=\frac{1}{1+z^2}}\,$ here is

$$Res_{z=i}(f)=\lim_{z\to i}(z-i)f(z)=\lim_{z\to i}\frac{1}{z+i}=\frac{1}{2i}$$

So by Cauchy's Residue Theorem we get:

$$\pi=2\pi i\frac{1}{2i}=\oint_\Gamma\frac{dz}{z^2+1}=\int_{-R}^R\frac{dx}{x^2+1}+\int_{\gamma_R}\frac{dz}{z^2+1}$$

But applying the evaluation theorem (by Cauchy, of course...who else?), we have

$$\left|\int_{\gamma_R}\frac{dz}{z^2+1}\right|\leq \max_{z\in\gamma_r}\left(\frac{1}{z^2+1}\right)\cdot \pi R\leq \frac{1}{R^2-1}\cdot \pi R\xrightarrow [R\to\infty]{} 0$$

So making $\,R\longrightarrow \infty\,$ ,we get:

$$\pi=\int_{-\infty}^\infty\frac{dx}{x^2+1}=2\int_0^\infty\frac{dx}{x^2+1}$$ as the function's clearly even.

Also note the simple definite Riemann integral $$\left.\int_0^1\frac{1}{x^2+1}=\arctan x\right|_0^1=\frac{\pi}{4}$$ so finally:

$$\frac{\pi}{2}=\int_0^\infty\frac{dx}{x^2+1}=\int_0^1\frac{dx}{x^2+1}+\int_1^\infty\frac{dx}{x^2+1}\Longrightarrow \int_1^\infty\frac{dx}{x^2+1}=\frac{\pi}{4}$$

  • $\begingroup$ Thanks for the answer, although it does not follow the steps that the statement requires. I have seen a similar exercise where it adds a logarithmic branch point in the lower limit (the complex function would be $F(z)=\frac{\log (z-1)}{1+z^2}$). Thus we have two circumferences, one with radius $\varepsilon$ and other with radius $R$, both with center in $z=1$. The branch cut would be $\log (x-1)+\mathrm{i}0-\log (x-1)+\mathrm{i}2\pi = -\mathrm{i}2\pi$ But I don't know exactly why it is solved this way and how to continue it. $\endgroup$ – Mike Jul 2 '12 at 0:55
  • $\begingroup$ I don't follow you, @Mike: one thing is the complex logarithm and another what you have. You don't need any branch stuff here as you give in your comment above, which has nothing to do with the question you asked. $\endgroup$ – DonAntonio Jul 2 '12 at 1:40
  • $\begingroup$ Well, the exact model exercice I am following in order to do the same steps that are required is $I=\int _0 ^1 \frac{\mathrm{d}x}{1+x^2}$. I know there is no need of branch points, but for some reason it is asked. Here they use the function $F(z)=\frac{\log (z^\alpha (z-1)^\beta)}{1+z^2}$. This way there are two branch points: in $z=0$ and in $z=1$. There are two simple poles from the denominator ($\pm \mathrm{i}$) and two singularities from the branch points ($0$, $1$). There'll be two contours, the circumference of center $0$ and radius $r$ and the one of center $1$ and radius $r$. $\endgroup$ – Mike Jul 2 '12 at 7:03
  • $\begingroup$ (Continuation) Then $\oint _\Gamma F(z) \mathrm{d}z=2\pi \mathrm{i} \sum \mathrm{Res} [F(z), \pm \mathrm{i}]=$ $=\int_0 ^1 \Delta_0 F(x) \mathrm{d}x + \oint _{\delta(1,\varepsilon)} F(z) \mathrm{d}z + \int_1 ^{\infty} \Delta_1 F(x) \mathrm{d}x + \oint_{\delta(1,r)} F(z) \mathrm{d}z + \oint_{\delta(0,\varepsilon)} F(z) \mathrm{d}z$ Also $\log z^\alpha =\alpha \log (x+\mathrm{i}0)$ and $\log(z-1)^\beta = \beta \log((x-1)+\mathrm{i}0)$. $\Delta_1 = -\frac{\mathrm{i}2\pi (\alpha+\beta)}{1+x^2}$ We don't want $\int_1 ^{\infty} \Delta_1 F(x) \mathrm{d}x$ to appear, so let be $\alpha + \beta=0$ $\endgroup$ – Mike Jul 2 '12 at 7:32
  • 1
    $\begingroup$ @Mike, unless your teacher wanted to make some rather focused point on branch points, complex logarithm or something like that, I honestly have no pale idea what would anyone do you real improper integral choosing the function $\,\frac{\log(z^\alpha(z-1)^\beta)}{1+z^2}\,$ . Perhaps this gives some rather important insights into your integral, but I can't see it. $\endgroup$ – DonAntonio Jul 2 '12 at 12:49

Well, I finally did it that way. Compute using complex methods: $$I=\int _1 ^\infty \frac{\mathrm{d}x}{x^2+1}$$

i) Choose the complex function to integrate. $$f(z)=\log (z-1) \frac{1}{z^2+1}$$ Singularities: $z_0 =1$ (branch point), $z_1 =\mathrm{i}$ and $z_2=-\mathrm{i}$ (simple poles).

ii) Choose the contour.

$\Gamma$ enclosed between two circumferences centred in $z=1$, $\delta(\varepsilon)$ and $\delta(R)$ with a branch cut between $0^+$ and $2\pi^-$. It contains the two poles.

iii) Compute the integrals along circumferences. $$\oint _{\delta(R)}f(z)\mathrm{d}z=\int _0 ^{2\pi} \log(1+R\mathrm{e}^{\mathrm{i}\theta}-1)\frac{\mathrm{i}R\mathrm{e}^{\mathrm{i}\theta}\mathrm{d}\theta}{(1+R\mathrm{e}^{\mathrm{i}\theta})^2+1}\xrightarrow [R\to\infty]{} 0$$

$$\oint _{\delta(\varepsilon)}f(z)\mathrm{d}z=\int _0 ^{2\pi} \log(1+\varepsilon\mathrm{e}^{\mathrm{i}\theta}-1)\frac{\mathrm{i}\varepsilon\mathrm{e}^{\mathrm{i}\theta}\mathrm{d}\theta}{(1+\varepsilon\mathrm{e}^{\mathrm{i}\theta})^2+1}\xrightarrow [\varepsilon\to 0]{} \varepsilon \log \varepsilon \rightarrow 0$$

iv) Compute the branch cut. $$\Delta=f\left( (x-1) \mathrm{e}^{\mathrm{i}0}\right) - \log\left( (x-1)\mathrm{e}^{\mathrm{i}2\pi}\right) = \frac{f\left( (x-1) \mathrm{e}^{\mathrm{i}0}\right) - \log\left( (x-1)\mathrm{e}^{\mathrm{i}2\pi}\right)}{x^2+1}=\frac{-2\pi\mathrm{i}}{x^2+1}$$

v) Compute the integral. $$I=\frac{2\pi\mathrm{i}}{\Delta}\sum\mathrm{Res}[f(z);\mathrm{i},-\mathrm{i}]=-\sum\mathrm{Res}[f(z);\mathrm{i},-\mathrm{i}]=-\left( \frac{\pi}{4} \right) = \frac{\pi}{4}$$

The residues at the poles are: $$\lim_{z\rightarrow\mathrm{i}}(z-\mathrm{i})\frac{\log(z-1)}{z^2+1}=\lim_{z\rightarrow\mathrm{i}}\frac{\log(z-1)}{z+\mathrm{i}}=\frac{\log(\mathrm{i}-1)}{2\mathrm{i}}$$ $$\lim_{z\rightarrow-\mathrm{i}}(z+\mathrm{i})\frac{\log(z-1)}{z^2+1}=\lim_{z\rightarrow-\mathrm{i}}\frac{\log(z-1)}{z-\mathrm{i}}=\frac{\log(-\mathrm{i}-1)}{-2\mathrm{i}}$$ $$\sum\mathrm{Res}[f(z);\mathrm{i},-\mathrm{i}]=\frac{\log(\mathrm{i}-1)-\log(-\mathrm{i}-1)}{2\mathrm{i}}=\frac{\log\frac{\mathrm{i}-1}{-\mathrm{i}-1}}{2\mathrm{i}}=\frac{-\mathrm{i}\frac{\pi}{2}}{2\mathrm{i}}=-\frac{\pi}{4}$$

vi) Compute the integral using elemental methods. $$I=\int_1 ^\infty \frac{\mathrm{d}x}{x^2+1}=\lim _{a\rightarrow \infty} \int _1 ^a \frac{\mathrm{d}x}{x^2+1}= \lim _{a\rightarrow \infty} \left[ \arctan x \right]_1 ^a =\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$


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