Improper integration using complex methods

Question

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).

Answer 1

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

Answer 2

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:

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

Answer 3

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.