# Really hard integrals?

pakmingki
can someone give me some really hard intergrals to solve?

make sure they are in the range of calculus 1-2 (anything before multivariable)

My teacher assigned some few hard integrals, and they are fun. I want to try moer.
thanks.

CollinsArg

yip
Try $$\int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}}$$

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itsjustme
sin(2x)cos(2x)dx

ObsessiveMathsFreak
$$\int e^{-x^2} dx$$

yucheng, ammarb32, CollinsArg and 3 others
Homework Helper
$$\int e^{-x^2} dx$$

I doubt that it belongs to either Calculus 1 or Calculus 2 problems.

pakmingki said:
... make sure they are in the range of calculus 1-2 (anything before multivariable)...

yucheng and Delta2
Homework Helper
$$\int^{1}_0 \frac{\log_e (1+x)}{x} dx$$. Quite an interesting one that someone gave to me. Nice Solution :)

Staff Emeritus
Gold Member
Find $$\frac{f'(x)}{f(x)}$$ where $$f(x) = sec(x)+tan(x)$$and hence find $$\int sec(x) dx$$.

One of my faves

pakmingki
wow, these loko pretty fun. THey look way different from the ones I've ever seen.

Ill give them a whirl sometime soon.

bob1182006
This is a pretty hard one but I haven't finished Calc 2 so I don't know any harder than this.

My favorite Integral so far is this:

$$\int \frac{dx}{(x^2+9)^3}$$

It's general form is of
$$\int \frac{dx}{(x^2+a^2)^n}$$

It has a really interesting answer

1 person
zoki85
Hard ,but famous and bautiful :

$$\int_{0}^{\infty}sin(x^2)dx$$

janhaa
Try this one...

$$\int \sqrt{\tan(x)}{\rm dx}$$

bit188
Try this one...

$$\int \sqrt{\tan(x)}{\rm dx}$$
That's a good one :rofl:

Invictious
Hard ,but famous and bautiful :

$$\int_{0}^{\infty}sin(x^2)dx$$

Took me 5 minutes only
That question though, however, was just..simply amazing.
I suggest everyone try that question

Homework Helper
I think the original poster has quite enough thanks...he hasn't actually done any of them yet.

JohnDuck
Hard ,but famous and bautiful :

$$\int_{0}^{\infty}sin(x^2)dx$$

I'm stumped but intrigued.

Equilibrium
$$\int \frac{1}{x^5+1}dx$$

zoki85
Took me 5 minutes only
That question though, however, was just..simply amazing.
I suggest everyone try that question

We are not all as clever as you Invictious :tongue:

yucheng
DyslexicHobo
$$\int \frac{1}{x^5+1}dx$$

I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

$$\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr$$

I used u-substitution (well, r-substitution), where $$r = x^5 + 1$$. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

And please tell us how to do $$\int_{0}^{\infty}sin(x^2)dx$$

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$$\int e^{-x^2} dx$$
how can this even be integrated?:uhh:

JohnDuck
how can this even be integrated?:uhh:

It can be proved that there's no elementary antiderivative, but you can use a trick from multivariable calculus involving a change to polar coordinates and the squeeze theorem to evaluate it. It's called a Gaussian integral.

Edit: Correction--the trick works for $$\int_{- \infty}^{\infty} e^{-x^{2}}dx$$

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janhaa
Integration

how can this even be integrated?:uhh:

Observe that $$\,e^{-x^2}\,$$is an even function, and we can integrate in two dimensions ):

$$I^2=(\int_{\mathbb{R}} e^{-x^2}){\rm dx})^2=(\int_{\mathbb{R}}e^{-x^2}{\rm dx})(\int_{\mathbb{R}}e^{-y^2}{\rm dy})$$

$$I^2=\int_{\mathbb{R}} \int_{\mathbb{R}} e^{-(x^2+y^2)}{\rm dx}{\rm dy}$$

Then change to polar coordinates:

$$I^2=\int_0^{2\pi}\int_0^{\infty} e^{-r^2} r {\rm dr} {\rm d\theta}=2\pi \int_0^{\infty} e^{-r^2} r {\rm dr}$$

then substitution:

$$\, u = r^2 \,$$

$$\frac{\rm du}{2r}={\rm dr}$$

that is:

$$I^2=\pi \int_0^{\infty} e^{-u} {\rm du}= \pi$$

finally:

$$I=\int_{\mathbb{R}} e^{-x^2} {\rm dx}=\int_{- \infty}^{\infty} e^{-x^2} {\rm dx}=\sqrt{\pi}$$

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Maarten Havinga
Staff Emeritus
Gold Member
Way to drop the ball on the limits at the end...

yip
I'm stumped but intrigued.

I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

$$\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr$$

I used u-substitution (well, r-substitution), where $$r = x^5 + 1$$. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

And please tell us how to do $$\int_{0}^{\infty}sin(x^2)dx$$
For that integral, i think u have to use the partial fraction theorem that involves derivatives, as there is trigonometric terms in the primitive of that function. Here is how I did the second integral, since u asked, the trick is to apply euler's theorem:

$$e^{-ix^{2}}=cosx^{2}-isinx^{2}$$
$$\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx$$
It is known that $$\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}$$
$$\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}$$
$$\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}$$
$$\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)$$
$$\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$

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Aritro Biswas
JohnDuck
For that integral, i think u have to use the partial fraction theorem that involves derivatives, as there is trigonometric terms in the primitive of that function. Here is how I did the second integral, since u asked, the trick is to apply euler's theorem:

$$e^{-ix^{2}}=cosx^{2}-isinx^{2}$$
$$\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx$$
It is known that $$\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}$$
$$\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}$$
$$\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}$$
$$\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)$$
I follow you up to here.
$$\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$
Wha?

yip
Its simply taking the imaginary part of the integral, as the imaginary part of e^-ix^2 is -sinx^2

JohnDuck
Oh. That makes sense.

DyslexicHobo
I followed you up to about the part where... uhh nevermind. Didn't catch any of that. :/

Way above my head. Thanks for the explanation, though. I don't even understand how we can even begin to integrate a transcendental function using limits of infinity. They don't have a value at infinity, so how can they be evaluated?

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JohnDuck
Improper integrals such as:

$$\int_{a}^{\infty} f(x)dx$$

are defined as such:

$$\lim_{b\rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{b\rightarrow \infty} F(b) - F(a)$$

where F(x) is the antiderivative of f(x). If you don't understand how to take limits of functions at infinity, you should probably read up on limits again. Most calculus texts have a brief section on limits, and any introductory real analysis text certainly covers the topic thoroughly.

FlashStorm
I am scared and frightened.

Staff Emeritus
Gold Member
I am scared and frightened.

:rofl::rofl:

DyslexicHobo
Improper integrals such as:

$$\int_{a}^{\infty} f(x)dx$$

are defined as such:

$$\lim_{b\rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{b\rightarrow \infty} F(b) - F(a)$$

where F(x) is the antiderivative of f(x). If you don't understand how to take limits of functions at infinity, you should probably read up on limits again. Most calculus texts have a brief section on limits, and any introductory real analysis text certainly covers the topic thoroughly.

Sorry, I must have mis-worded myself. I meant that I don't understand how trigonometric functions can be evaluated as their inside approaches infinity. The function oscillates between 1 and -1, and never converges. I'm assuming that the Fundamental Theorem cannot be used here befause $$\lim_{x\rightarrow\infty} 2xcos(x^2)$$ (2xcos(x^2) is the anti-derivative of the starting function) cannot be evaluated, so it seems.

JohnDuck
Sorry, I must have mis-worded myself. I meant that I don't understand how trigonometric functions can be evaluated as their inside approaches infinity. The function oscillates between 1 and -1, and never converges. I'm assuming that the Fundamental Theorem cannot be used here befause $$\lim_{x\rightarrow\infty} 2xcos(x^2)$$ (2xcos(x^2) is the anti-derivative of the starting function) cannot be evaluated, so it seems.

That's actually not the antiderivative of $\sin{x^{2}}$. You're right in that, for example, $\lim_{x\rightarrow \infty} \sin{x}$ doesn't converge in the real numbers. This can be proven pretty easily. However, depending on the function inside of sine, it may converge. Consider $\lim_{x\rightarrow \infty} \sin{\frac{1}{x}}$--it converges to 0.

Edit: My gut says that $\sin{x^{2}}$ has no elementary antiderivative, but I'm not sure how one might prove or disprove this.

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Homework Helper
Well by integrating the Taylor series of sin(x^2) term by term we get:

$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{4n-1}}{(4n-1)(2n+1)!}$$

which one could recognize as $$\sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)$$

where $$S(u)=\int^u_0 \sin \left(\frac{1}{2} \pi x^2\right) dx$$.

As S(u) is not an elementary function, we have proved $\sin x^2$ has no elementary derivative.

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Aritro Biswas
yip
Well by integrating the Taylor series of sin(x^2) term by term we get:

$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{4n-1}}{(4n-1)(2n+1)!}$$

which one could recognize as $$\sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)$$

where $$S(u)=\int^u_0 \sin \left(\frac{1}{2} \pi x^2\right) dx$$.

As S(u) is not an elementary function, we have proved $\sin x^2$ has no elementary anti-derivative.

Isn't that a sort of circular argument? If u assume S(u)=integral of sin(cx^2), where c=pi/2 does not have a closed form answer, aren't u also implicitly assuming that the integral of sin(x^2), the case where c=1, also does not have a closed form answer? I was under the impression that changing constants doesn't really affect integrability, only changing the variables that u are integrating with respect to would. I think that proving that something is not integratable would involve much more complicated arguments. This is just my opinion though. My attempt at saying something is not integrable would rely on the well known fact that e^-ix^2 has no closed form integral, and since the sum of the integral of isin(x^2) and cos(x^2) equals the integral of e^-ix^2, if a closed form integral did exist for sin(x^2), then one must also exist for cos(x^2) (since the cosine function is just a shifted sine function), which means a closed form integral exists for e^-ix^2, which is a contradiction.

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Homework Helper
Thats what a lot of non-elementary functions are :) Functions invented with the pure purpose of being the antiderivative of something that otherwise wouldn't have one. eg The Error function, of SI(x)