Solutions check. Find $\int_{-b}^{b} \Big( \frac{1}{t + ix} - \frac{1}{t - ix} \Big) dt$

Question

Let $x > 0$. Find $$\lim_{b \to \infty} \int_{-b}^{b} \Big( \frac{1}{t + ix} - \frac{1}{t - ix} \Big) dt$$

I solved this problem but I am not quite confident about my steps. So please, could you check my solutions and tell me if there is something wrong??. Thanks in advanced!

First approuch:

Let's just focus on the first member of the integral: $$\int_{-b}^{b} \frac{1}{t + ix} dt = \int_{-b}^{b} \frac{1}{t + ix} \Big( \frac{t - ix}{t - ix} \Big) dt = \int_{-b}^{b} \frac{t - ix}{t^2 + x^2} dt$$

where the integrand can be again separated:

$$ = \int_{-b}^{b} \frac{t}{t^2 + x^2}dt - \int_{-b}^{b} \frac{ix}{t^2 + x^2}dt$$

Then, since the integral is about a real number $t$, $i$ can be treated as a constant. Now, I notice that the second integrand is an even function so the integral from $-b$ to $b$ is $0$. So we have:

$$ = \int_{-b}^{b} \frac{t}{t^2 + x^2}dt = \frac{1}{2}\ln\Big(t^2 + x^2 \Big) \Big\rvert_{-b}^{b} = \frac{1}{2}\ln\Big(\frac{b^2 + x^2}{b^2 + x^2} \Big) = 0$$

Following almost the same steps one can also show that $$\int_{-b}^{b} \frac{1}{t - ix} dt= 0$$

So the question reminds as $$\lim_{b \to \infty} (0 + 0) = 0$$

Is this correct?? I mean:

• Can $i$ be really treated as a constat in an integral which is about a real number??

• Why would the question ask for about a limit int the infinite if the integral was destined to be $0$??

Second approuch

If $i$ can be really treated as a constant then the integral of the function $\frac{1}{t + ix}$ would be $\ln(t + ix)$, since $ix$ would be also a constant, so we have:

$$\int_{-b}^{b} \Big( \frac{1}{t + ix} - \frac{1}{t - ix} \Big) dt= \ln(t + ix)\Big\rvert_{-b}^{b} - \ln(t - ix)\Big\rvert_{-b}^{b}$$

Since $ix$ is just a constant then $\ln$ fullfills the same properties as the well-known real valued function $\ln$, so:

$$ = \ln\Big(\frac{b + ix}{-b + ix} \Big) - \ln\Big(\frac{b - ix}{-b - ix} \Big) = \ln\Big(\frac{\frac{b + ix}{-b + ix}}{\frac{b - ix}{-b - ix}} \Big) = \ln(1) = 0$$

Soo the answer would be $\lim_{b \to \infty} 0 = 0$. I am really confused with theese steps since $\ln$ function was evaluated in $b + ix$ and although I know that there is a counter-part complex version of the real function $\ln$, I had accepted $ix$ as a costant, so there wouldn't be nothing wrong, right??

Third approuch

If $i$ can be really treated as a constant, then:

$$\int_{-b}^{b} \Big( \frac{1}{t+ ix} - \frac{1}{t - ix} \Big) dt = \int_{-b}^{b} \Big( \frac{-2ix}{t^2 + x^2} \Big) dt$$

because the integrand is an even function, that integral is $0$. So we have again $\lim_{b \to \infty} 0 = 0$

Any comment about what I wrote would be really appreciated!

Answer 1

If $x=0$, then the integral vanishes. Assume that $x\ne 0$. Then for $x>0$ $$ \int_{-b}^b\left(\frac{1}{t+ix}-\frac{1}{t-ix}\right)\,dt= \int_{-b}^b\frac{-2ix\,dt}{t^2+x^2}=\int_{-b}^b\frac{-2i\,d(t/x)}{(t/x)^2+1} =-2i\int_{-b/x}^{b/x}\frac{ds}{s^2+1}=-2i \big(\tan^{-1}(b/x)-\tan^{-1}(-b/x)\big)=-4i\tan^{-1}(b/x)\to -4i\cdot\frac{\pi}{2}=-2\pi i, $$ For $x<0$, the limit is $2\pi i$.

Answer 2

The problem with your first and third approaches is that you wrote that any even function $f(x)$, we get that $\int_{-b}^b f(x) dx=0$. This statement is true for odd functions. For even functions, $\int_{-b}^b f(x) dx=2\int_0^b f(x) dx$.

The problem with your second approach is that in the world of complex functions, logarithms are multiple-valued, and as such give you an answer that is true up to a $+2k\pi i, k\in\mathbb{Z}$.