Tricks for taking complex integrals

Integrals, what can be better? Well, perhaps not for everyone, but all the same, I have not lost anything so strictly in mathematic, so much as anything else. This is the way - how to get "complex" intrals. This is reasoning for what the reader has learned in the school and knows the trivial materialities (for example, integration in parts ). In the future, we will discuss only the Roman Intermediate, but not the Intermediate Lebes-Styles, Ito, Skoroksoda, and so on (but I would be pleased with everything else,).

Take this - a small selection of recipes or “patter” drivers that you can take into the pile and use it. Then read the high-DIP format in order to turn the eye on the eye. I prudently.

Transition to Polar Coordinates

From a small hackneyed method - a transfer to the Polar coordinators. It is noteworthy that the transfer to the plots of the co-ordinates can be used even in those places, it would be better to say that of the co-ordinates of the co-ordinates. Supplement, indefinite Gauss integral $\ textstyle \ int e ^ {- x ^ 2} {\ mathrm d} x$ have no alternative decision, and now a definite integer $\ textstyle \ int _ {- \ infty} ^ {\ infty} e ^ {- x ^ 2} {\ mathrm d} x = \ sqrt {\ pi}$ .

To prove it as possible: as a start, in order to take advantage of the transfer, we introduce two variable conversions. $\ textstyle x$ and $\ textstyle y$ so what

$I = \ int _ {- \ infty} ^ {\ infty} e ^ {- x ^ 2} {\ mathrm d} x = \ int _ {- \ infty} ^ {\ infty} e ^ {- y ^ 2} { \ mathrm d} y$

Devices coordinates can be expressed through the polar $\ textstyle (r, \ theta)$ hereby so

$\ begin {align *} x & amp; = r \ cos \ theta \\ y & amp; = r \ sin \ theta \\ r ^ 2 & amp; x ^ 2 + y ^ 2 \ end {align *}$

Integration from $\ textstyle - \ infty$ to $\ textstyle \ infty$ in the coordinate system - this is what the integration $\ textstyle r$ from $\ textstyle 0$ to $\ textstyle \ infty$ and $\ textstyle \ theta$ from $\ textstyle 0$ to $\ textstyle 2 \ pi$ .

We get the following results:

$\ begin {aligned} I \ cdot I & amp; = \ int _ {- \ infty} ^ {\ infty} e ^ {- x ^ 2} {\ mathrm d} x \ int _ {- \ infty} ^ {\ infty} e ^ {- y ^ 2} {\ mathrm d} y \\ & amp; = \ int _ {- \ infty} ^ {\ infty} \ int _ {- \ infty ^ {\ infty} e ^ {- x ^ 2 -y ^ 2} \; {\ mathrm d} x \; {\ mathrm d} y \\ & amp; = \ int_ {0} ^ {2 \ pi} {\ mathrm d} \ theta \ int_ {0} ^ {\ infty} e ^ {- r ^ 2} r \; {\ mathrm d} r \\ & amp; = 2 \ pi \ int_ {0} ^ {\ infty} e ^ {- r ^ 2} r \; {\ mathrm d} r \\ & amp; = \ pi \ int_0 ^ {\ infty} e ^ {- r ^ 2} \; {\ mathrm d} r ^ 2 = \ pi \\ \ end {aligned}$

$\ therefore I = \ sqrt {\ pi}$

This approach can also be used in 3 measurements with the use of the physical coordinates. $\ textstyle (x, y, z) \ rightarrow (r, \ theta, \ phi)$ .

GEOMETRIC INTERPRETATION

Boost, “rolling in geometry,” fetches food. Boot, for example, let us count you.

$\ int_0 ^ \ infty \ frac {{\ mathrm d} x} {1 + x ^ 2}$

I am sure that many of you know that there is an analytical solution for this integrator. $\ textstyle \ tan ^ {- 1} x$ , therefore, counting a definite intranet is not a problem. But on the other hand, this is an integral, you can even count without this knowledge.

Submit a circle to the radio $\ textstyle r$ with the center $\ textstyle (0,0)$ . The length of the arc of this circle with a central corner $\ textstyle \ theta$ paved $\ textstyle L = r \ theta$ , and if the circle is one - just $\ textstyle \ theta$ . Togda

$L = \ theta = \ int_0 ^ {\ theta} \; {\ mathrm d} t$

where $\ textstyle t$ - this is a repetitive integration.

In this way, the subgroup expression is equal to $\ textstyle 1$ but we can fortify him, for example

$\ begin {align *} L & amp; = \ int_0 ^ {\ theta} 1 \; {\ mathrm d} t \\ & amp; = \ int_0 ^ {\ theta} \ frac {\ frac {1} {\ cos ^ 2t}} {\ frac {1} {\ cos ^ 2t}} \; {\ mathrm d} t \\ & amp; = \ int_0 ^ {\ theta} \ frac {\ frac {1} {\ cos ^ 2t} } {\ frac {\ cos ^ 2t + \ sin ^ 2t} {\ cos ^ 2t}} \; {\ mathrm d} t \\ & amp; = \ int_0 ^ {\ theta} \ frac {\ frac {1} { \ cos ^ 2t}} {1+ \ tan ^ 2t} \; {\ mathrm d} t \\ \ end {align *}$

Next, let's make a submission.

$x = \ tan t \ Rightarrow {\ mathrm d} x = \ frac {{\ mathrm d} t} {\ cos ^ 2 t}$

By the amount we receive

$L = \ int_0 ^ {\ tan \ theta} \ frac {{\ mathrm d} x} {1 + x ^ 2}$

Let's get what $\ textstyle \ theta = \ frac {\ pi} {2}$ . Togda $\ textstyle \ tan \ theta = \ tan \ frac {\ pi} {2} = \ infty$ a little $\ textstyle \ frac {\ pi} {2}$ dies on the vertex of the circle (the length of the whole circle) $\ textstyle 2 \ pi$ ), we will instantly get a result

$\ frac {\ pi} {2} = \ int_0 ^ {\ infty} \ frac {{\ mathrm d} x} {1 + x ^ 2}$

By analogy with this result, you can also get other, breaking the circle differently, for example,

$\ begin {align *} \ frac {\ pi} {4} & amp; = \ int_0 ^ 1 \ frac {{\ mathrm d} x} {1 + x ^ 2} \\ \ frac {\ pi} {3} & amp; = \ int_0 ^ {\ sqrt {3}} \ frac {{\ mathrm d} x} {1 + x ^ 2} \\ \ end {align *}$

and so on.

Splitting diapason of integration

Let us count you

$\ int_0 ^ {\ infty} \ frac {\ ln x} {1 + x ^ 2} \; {\ mathrm d} x$

For the taking of this interpolation, the intrusion of the integration of the two, because $\ textstyle \ int_0 ^ {\ infty} = \ int_0 ^ 1 + \ int_1 ^ {\ infty}$ .

Borrowed by the first interlocutor, i.e. $\ textstyle \ int_0 ^ 1$ . Let's make a submission $\ textstyle t = 1 / x \ Rightarrow {\ mathrm d} x = - {\ mathrm d} t / t ^ 2$ . Get

$\ begin {align *} \ int_0 ^ 1 \ frac {\ ln x} {1 + x ^ 2} \; {\ mathrm d} x & amp; = \ int _ {\ infty} ^ 1 \ frac {\ ln (1 / t)} {1 + 1 / (t ^ 2)} \ left (- \ frac {1} {t ^ 2} \; {\ mathrm d} t \ right) \\ & amp; = - \ int _ {\ infty} ^ 1 \ frac {\ ln (1 / t)} {t ^ 2 + 1} \; {\ mathrm d} t \\ & amp; = \ int_1 ^ {\ infty} \ frac {\ ln (1 / t)} {t ^ 2 + 1} \; {\ mathrm d} t \\ & amp; = - \ int_1 ^ {\ infty} \ frac {\ ln t} {t ^ 2 + 1} \; {\ mathrm d} t \ end {align *}$

To externally rendered, which is a variable $\ textstyle t$ does the same function $\ textstyle x$ . Other words $\ textstyle \ int_0 ^ 1 = - \ int_1 ^ {\ infty}$ And this means that we automatically derive the value of a suitable integrator:

$\ int_0 ^ {\ infty} \ frac {\ ln x} {1 + x ^ 2} \; {\ mathrm d} x = 0$

Split between the honor and the insensitive

Not necessary for you, for example, to count

$\ int _ {- 1} ^ {1} \ frac {\ cos x} {e ^ {1 / x} +1} \; {\ mathrm d} x$

Let's make some substitutions:

$\ begin {align *} f (x) & amp;: = e ^ {1 / x} \\ g (x) & amp; = = \ frac {\ cos x} {f (x) +1} \ end {align *}$

Now we need to count $\ textstyle \ int _ {- 1} ^ {1} g (x) \; {\ mathrm d} x$ , and here the most interesting is begun. We celebrate $\ textstyle g (x)$ as the sum of honor and odd functions:

$g (x) = g_e (x) + g_o (x)$

Many ask for “but what can we do in the first place?” - on the one hand, yeah, and in a blessing. Understanding and comprehending above $\ textstyle -x$ around the world $\ textstyle x$ . You will get

$g (-x) = g_e (-x) + g_o (-x) = g_e (x) - g_o (x)$

no less than the integrity and inaccuracy of functions. In addition, we can express the interesting and odd side of the function as

$g_e (x) = \ frac {g (x) + g (-x)} {2}$

and

$g_o (x) = \ frac {g (x) -g (-x)} {2}$

So Cooperative, our intros can be reputed as

$\ int _ {- 1} ^ {1} g (x) \; {\ mathrm d} x = \ int _ {- 1} ^ {1} g_e (x) \; {\ mathrm d} x + \ int _ {- 1} ^ {1} g_o (x) \; {\ mathrm d} x = \ int _ {- 1} ^ {1} g_e (x) \; {\ mathrm d} x$

As you can see above, the inadequate function of the fray is full, there is only a four-minute compass, because

$\ int _ {- 1} ^ {1} g_o (x) \; {\ mathrm d} x = 0$

Okay, you have already set up the hope of waiting for this example. So here we have a formula $\ textstyle g_e (x) = \ frac {g (x) + g (-x)} {2}$ Give feedback to this form $\ textstyle g (x)$ . We will get

$g_e (x) = \ frac {1} {2} \ left (\ frac {\ cos x} {f (x) +1} + \ frac {\ cos (-x)} {f (-x) +1 } \ right)$

Ho we know what $\ textstyle \ cos x$ - fourth function, therefore $\ textstyle g_e (x)$ You can spell it like

$\ begin {align *} g_e (x) & amp; = \ frac {\ cos x} {2} \ left (\ frac {1} {f (x) +1} + \ frac {1} {f (-x ) +1} \ right) \\ & amp; = \ frac {\ cos x} {2} \ left (\ frac {f (-x) + 1 + f (x) +1} {f (x) f ( -x) + f (x) + f (-x) +1} \ right) \\ & amp; = \ frac {\ cos x} {2} \ left (\ frac {2 + f (-x) + f (x)} {f (x) f (-x) + f (x) + f (-x) +1} \ right) \\ \ end {align *}$

It is as it is and it is not real what is to be done with it. But take care of each other, and have a pause at us in the Formula. $\ textstyle f (x) f (-x)$ . Let us remember that $\ textstyle f (x) = e ^ {1 / x}$ and we will get

$f (x) f (-x) = e ^ {1 / x} e ^ {- 1 / x} = e ^ 0 = 1$

Well, that's all - our creeping stuff above our compatriot the number and denominator are equal, and this means something

$g_e (x) = \ frac {\ cos x} {2}$

And I’m easy to calculate:

$\ begin {align *} \ int _ {- 1} ^ {1} \ frac {\ cos x} {e ^ {1 / x} +1} \; {\ mathrm d} x & = \ int _ {- 1 } ^ {1} \ frac {\ cos x} {2} \; {\ mathrm d} x = \ sin (1) = 0.841 ... \ end {align *}$

Want more?

I have come to understand in what way, for the sake of one day, quite a bit. If I didn’t write so well, I read the mathematics of zero books (what is yours and my own), so that term crime might be straggling.

There is another Vagon of different stupas, so, if it is an interesting, glorious newsgroup. Udachi! ■

Source: https://habr.com/ru/post/314820/

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